Answer:
The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m
Explanation:
Worked : work can be defined as the product of force and distance.
The S.I unit of work is Joules (J).
Mathematically it can be represented as,
W = F×d.................. Equation 1
d = W/F.............................. Equation 2
where W = work, F = force, d = distance.
Given: W = 12 J
(i) for the 3.0 N weight,
using equation 2
d = 12/3
d= 4 m.
(ii) for the 4.0 N weight,
d = 12/4
d = 3 m.
(iii) for the 6.0 N weight,
d = 12/6
d = 2 m.
(iv) for the 2.0 N weight,
d = 12/2
d = 6 m
Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m
A 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s. (a) Why does the tackle constitute a perfectly inelastic collision? (b) Calculate the velocity of the players immediately after the tackle and (c) determine the mechanical energy that is lost as a result of the collision. (d) Where did the lost energy go?
Answer:
a) Please see below as the answer is self-explanatory.
b) 2.88 m/s
c) 785. 8 J
d) It is expended like thermal energy, due to internal friction.
Explanation:
a) In a tackle, both players keep emmeshed each other, so it is a perfectly inelastic collision; Immediately after the tackle, both masses behave like they were only one.
b) Assuming no external forces act during the collision, total momentum must be conserved.
As momentum is a vector, the conservation principle must be met by all vector components at the same time.
In our case, as the players move in directions mutually perpendicular, we can decompose the momentum vector along both directions, taking into account that after the collision, the momentum vector will have components along both directions.
So, if we call the W-E axis our X-axis (being the direction towards east as the positive one) , and to the S-N axis our Y -axis (being the northward direction the positive one), we can write the following equations:
pₓ₀ = pₓf ⇒ m₁*v₁ = (m₁+m₂)*vf*cosθ
py₀ = pyf ⇒ m₂*v₂ = (m₁+m₂)*vf*sin θ
where θ, is the angle that both players take regarding the x-axis after the collision (north of east).
Replacing by the values, we have the following equations:
vf*cosθ = (90.0 kg*5.00 m/s) / (90.0 kg + 95.0 kg) = 2.43 m/s (1)
vf*sin θ = (95.0 kg* 3.00 m/s) / (90.0 kg + 95.0 kg) = 1.54 m/s (2)
Dividing both sides:
sin θ / cos θ = tan θ = 1.54 / 2.43 = 0.634
⇒ arc tan (0.634) = 32.3º
Replacing in (1) we have:
vf = 2.43 m/s / cos 32.3º = 2.43 m/s / 0.845 = 2.88 m/s
c) As the collision happens in one dimension, all mechanical energy, before and after the collision, is just the kinetic energy of the players.
Before the collision:
K₀ = 1/2*m₁*v₁₀² + 1/2 m₂*v₂₀²
= 1/2*( ( 90.0) kg*(5.0)²(m/s)² + (95.0)kg*(3.0)(m/s)²) = 1,553 J
After the collision:
Kf = 1/2 *(m₁+ 767.2 Jm₂)*vf² = 1/2*185 kg*(2.88)²(m/s)²= 767.2 J
The mechanical energy lost during the collision is just the difference between the final and initial kinetic energy:
ΔK = Kf - K₀ = 767.2 - 1,553 J = -785.8 J
So, the magnitude of the energy lost during the collision is 785.8 J.
d) This energy is lost during the collision as thermal energy, due to the internal friction between both players.
The tackle constitutes a perfectly inelastic collision where the players stick together after the collision, resulting in a loss of kinetic energy. The velocity of the players immediately after the tackle is 2.70 m/s to the east. The mechanical energy lost as a result of the collision is 562.5 J.
Explanation:(a) The tackle constitutes a perfectly inelastic collision because the two players stick together after the collision, resulting in a loss of kinetic energy. In a perfectly inelastic collision, the objects involved stick together and move as a single unit.
(b) To calculate the velocity of the players immediately after the tackle, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision. Since the fullback is running east, we can consider the positive direction as east and the negative direction as north. Applying the principle of conservation of momentum in the x-direction, we have:
Total momentum before the collision in the x-direction: (90.0 kg)(5.00 m/s) = 450 kg·m/sTotal momentum after the collision in the x-direction: (90.0 kg + 95.0 kg) * Vx = (185.0 kg) * VxSetting the two equations equal to each other and solving for Vx, we get:
(90.0 kg)(5.00 m/s) = (185.0 kg) * VxVx = 2.70 m/sSo the velocity of the players immediately after the tackle is 2.70 m/s to the east.
(c) The mechanical energy that is lost as a result of the collision can be calculated by subtracting the final kinetic energy from the initial kinetic energy. The initial kinetic energy is given by:
Initial kinetic energy = 0.5 * (90.0 kg) * (5.00 m/s)^2 = 562.5 JSince the players come to rest after the collision, the final kinetic energy is zero. Therefore, the mechanical energy lost is equal to the initial kinetic energy:
Mechanical energy lost = Initial kinetic energy = 562.5 J(d) The lost energy is converted into other forms of energy, such as sound, heat, and deformation of the players and their surroundings.
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While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 17m , and you use your watch to find that each loop around takes 26sWhat is your speed?Express your answer to two significant figures and include the appropriate units.What is the magnitude of your acceleration?Express your answer to two significant figures and include the appropriate units.What is the ratio of your weight at the top of the ride to your weight while standing on the ground?Express your answer using two significant figures.
Answer:
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
Explanation:
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed theCase Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
Which of the following statements are true? A. Earth's gravity has no effect on astronauts inside the International Space Station. B. An astronaut's mass is greater on Earth than on the Moon. C. An astronaut's weight is the same on the Moon as on Earth. D. An astronaut's mass is the same on the International Space Station as it is on Earth. E. None of these statements are true.
The given statement "An astronaut's mass is the same on the International Space Station as it is on Earth" is true.
Answer: Option D
Explanation:
There is usually a slight difference between mass and the weight of an object. The difference is that the mass of any object is independent of its acceleration due to gravity or gravitational influence of the planet where it is present.
Similarly, the weight of any object will be influenced by the gravitational force of that planet as the weight is directly proportional to the acceleration due to gravity of that planet.
So, the other three options are false and those three options states that weight of an object on Earth is equal to the weight of that object on any other planet. This is not true. So, the fourth option related to the mass of an astronaut in and outside Earth is true as it is equal theoretically.
An astronaut's mass being the same on the International Space Station as it is on Earth is a true statement.
What is Mass?Thus is defined as the resistance a matter offers to a change in its speed or position when force is applied.
Gravitational force doesn't determine the mass of objects which is why an astronaut's mass on International Space Station will be the same as on Earth. This therefore makes option D the most appropriate choice.
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We go out to sunbathe on a warm summer day. If we soak up 80 British thermal units per hour [BTU/h] of energy, how much will the temperature of 65 comma 000-gram person increase in 2 hours [h] in units of degrees Celsius [°C]? We assume that since our bodies are mostly water they have the same specific heat as water (4.18 joules per gram degree Celsius [J/(g degrees Upper C)]).
Answer:
0.62127°C
Explanation:
[tex]1\ BTU=1055\ J[/tex]
[tex]80\ BTU/h=80\times 1055=84400\ J/h[/tex]
Heat absorbed by body in 2 hours
[tex]Q=84400\times 2\\\Rightarrow Q=168800\ J[/tex]
m = Mass of person = 65000 g
c = Specific heat of water = 4180 J/kg°C
[tex]\Delta T[/tex] = Change in temperature
Heat is given by
[tex]Q=mc\Delta T\\\Rightarrow 168800=65\times 4180\times \Delta T\\\Rightarrow \Delta T=\dfrac{168800}{65\times 4180}\\\Rightarrow \Delta T=0.62127\ ^{\circ}C[/tex]
The increase in temperature will be 0.62127°C
A cylinder, which is in a horizontal position, contains an unknown noble gas at 4.00 × 10 4 Pa 4.00×104 Pa and is sealed with a massless piston. The piston is slowly, isobarically moved inward 16.3 cm, 16.3 cm, which removes 1.50 × 10 4 J 1.50×104 J of heat from the gas. If the piston has a radius of 30.5 cm, 30.5 cm, calculate the change in the internal energy of the system Δ U ΔU .
Answer:
-13094.55179 J
Explanation:
Q = Heat = [tex]-1.5\times 10^{4}\ J[/tex]
P = Pressure = [tex]4\times 10^4\ Pa[/tex]
[tex]\Delta V[/tex] = Change in volume = [tex]\pi r^2\times -h[/tex](negative because it is decreasing)
h = Height = 16.3 cm
r = Radius = 30.5 cm
Entropy is given by
[tex]\Delta U=Q-W[/tex]
Work done is given by
[tex]W=P\Delta V\\\Rightarrow W=4\times 10^4\times (\pi 0.305^2\times -0.163)[/tex]
[tex]\Delta U=-1.5\times 10^{4}-(4\times 10^4\times (\pi 0.305^2\times -0.163))\\\Rightarrow \Delta U=-13094.55179\ J[/tex]
The change in the internal energy of the system is -13094.55179 J
If the coefficient of static friction between tires and pavement is 0.60, calculate the minimum torque that must be applied to the 69-cm-diameter tire of a 920-kg automobile in order to "lay rubber" (make the wheels spin, slipping as the car accelerates). Assume each wheel supports an equal share of the weight.
To solve this problem, it is necessary to apply the definitions and concepts related to Newton's second law, which relate the variables of the Normal Force, Weight, friction force and finally the Torque.
We start under the definition that the Normal Force of one of the 4 tires of the car would be subject to
[tex]N = \frac{mg}{4}[/tex]
Where,
m = mass
g = Gravitational Acceleration
Therefore the Normal Force of each wheel would be
[tex]N = \frac{920*9.8}{4}[/tex]
[tex]N = 2254N[/tex]
Now the friction force can be determined as
[tex]f_s = \mu_s N[/tex]
[tex]f_s = 0.60 * 2254[/tex]
[tex]f_s = 1352.4N[/tex]
The radius of each of the tires is given as
[tex]r = \frac{69}{2}[/tex]
[tex]r = 34.5cm = 0.345m[/tex]
Finally, the torque is made between the friction force (which is to be overcome) and the radius of each of the wheels, therefore:
[tex]\tau = r*f_s[/tex]
[tex]\tau = (0.345)(1352.4)[/tex]
[tex]\tau = 466.578N\cdot m[/tex]
Therefore the engine of the car must apply a torque of about [tex]466.578N\cdot m[/tex] to lay rubber
To calculate the minimum torque needed to make the wheels spin on a car, we must first understand the concept of static friction. The force of static friction (Fs) that must be overcome to cause slipping is given by
Fs = μsN
where μs is the coefficient of static friction and N is the normal force. In this case, the weight of the car (W) is evenly distributed on all four tires, so each tire supports a quarter of the weight, W/4. The normal force N for one tire would then be W/4.
Since the weight W of the car is the mass (m) times the acceleration due to gravity (g), we have:
N = W/4 = mg/4.
Substituting the given values, we find
N = (920 kg * 9.81 m/s2)/4.
Using the coefficient of static friction (μs = 0.60), the static frictional force Fs for one tire is
Fs = 0.60 * N.
To find the torque (τ), we use the relation
τ = Fsr
where r is the radius of the tire.
The radius is half the diameter, so r = 69 cm / 2 or 0.345 m. Thus, the minimum torque is
τ = Fs * 0.345 m.
Calculating N, we get
N = (920 kg * 9.81 m/s2)/4
N = 2251.05 N
so Fs = 0.60 * 2251.05 N
Fs = 1350.63 N.
Therefore, the minimum torque τ is 1350.63 N * 0.345 m = 465.97 Nm.
Suppose that you release a small ball from rest at a depth of 0.600 m below the surface in a pool of water. If the density of the ball is 0.300 that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? Neglect any transfer of energy to the splashing and waves produced by the emerging ball.
Answer:
1.4m
Explanation:
A particle moves according to the law of motion s(t) = t^{3}-8t^{2}+2t, t\ge 0, where t is measured in seconds and s in feet. a.) Find the velocity at time t. Answer: b.) What is the velocity after 3 seconds? Answer: c.) When is the particle at rest? Enter your answer as a comma separated list. Enter None if the particle is never at rest. At t_1= and t_2= with t_1
Answer:
Explanation:
Given
displacement is given by
[tex]s(t)=t^3-8t^2+2t[/tex]
so velocity is given by
[tex]v(t)=\frac{\mathrm{d} s(t)}{\mathrm{d} t}[/tex]
[tex]v(t)=3t^2-16t+2[/tex]
(b)velocity after [tex]t=3 s[/tex]
[tex]v(3)=3(3)^2-8\cdot 3+2[/tex]
[tex]v(3)=19 m/s[/tex]
(c)Particle is at rest
when its velocity will become zero
[tex]v(t)=0[/tex]
i.e. [tex]3t^2-16t+2=0[/tex]
[tex]t=\frac{16\pm \sqrt{16^2-4\cdot 3\cdot 2}}{2\cdot 3}[/tex]
[tex]t=\frac{16\pm 15.23}{6}[/tex]
[tex]t=5.20 s[/tex]
Snorkeling by humans and elephants. When a person snorkels, the lungs are connected directly to the atmosphere through the snorkel tube and thus are at atmospheric pressure. In atmospheres, what is the difference Δp between this internal air pressure and the water pressure against the body if the length of the snorkel tube is
(a)24 cm (standard situation) and
(b)4.1 m (probably lethal situation)?
Answer:
2354.4 Pa
40221 Pa
Explanation:
[tex]\rho[/tex] = Density = 1000 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
h = Depth
The pressure difference would be
[tex]\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 0.24\times 9.81\\\Rightarrow \Delta P=2354.4\ Pa[/tex]
The pressure difference in the first case is 2354.4 Pa
[tex]\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 4.1\times 9.81\\\Rightarrow \Delta P=40221\ Pa[/tex]
The pressure difference in the second case is 40221 Pa
A heat engine takes thermal energy QH from a hot reservoir and uses part of this energy to perform work W. Assuming that QH cannot be changed, how can the efficiency of the engine be improved?
a) Increase the work W, the thermal energy QC rejected to the cold reservoir increasing as a result.
b) Decrease the work W, the rejected QC decreasing as a result.
c) Increase the work W, the rejected QC remaining unchanged.
d) Increase the work W, the rejected QC decreasing as a result.
e) Decrease the work W, the rejected QC remaining unchanged
Answer:
d) Increase the work W, the rejected QC decreasing as a result.
Explanation:
By the second law of thermodynamics the efficiency of a heat engine working between two reservoirs is:
[tex]\eta=\frac{W}{Q_{H}} [/tex] (1)
With W the work and [tex] Q_{H} [/tex] the heat of the hot reservoir, note in (1) that efficiency is directly proportional to the work and inversely proportional to the heat of the hot reservoir, so if we remain [tex]Q_{H} [/tex] constant we should increase the work to increase the efficiency.
Also, efficiency is:
[tex] \eta=1-\frac{Q_{C}}{Q_{H}}[/tex] (2)
With [tex]Q_{C} [/tex] the heat released to the cold reservoir, it is important to note that because second law of thermodynamics the efficiency of a heat engine should be between 0 and 1 ([tex]0\leq\eta\leq1 [/tex]), so the ratio [tex]\frac{Q_{C}}{Q_{H}} [/tex] always is positive and its maximum value is 1, that implies if [tex]Q_{H} [/tex] remains constant and efficiency increases, [tex]Q_{C} [/tex] will decrease and the ratio [tex]\frac{Q_{C}}{Q_{H}} [/tex] too.
So, the correct answer is d)
Final answer:
The efficiency of a heat engine is increased by increasing the work output while decreasing the rejected heat to the cold reservoir, in accordance with the first and second laws of thermodynamics.
Explanation:
To improve the efficiency of a heat engine, one must increase the work output, W, while simultaneously decreasing the heat rejected to the cold reservoir, Qc. The correct choice is:
d) Increase the work W, the rejected Qc decreasing as a result.
The efficiency, η, of a heat engine is defined as the ratio of the work done, W, to the heat absorbed from the hot reservoir, QH. By the first law of thermodynamics, QH = W + Qc, meaning that the efficiency can be increased either by increasing W or decreasing Qc. According to the second law of thermodynamics, there is a minimum amount of QH that cannot be used for work and must be rejected as Qc. Therefore, the aim is to minimize this rejected heat without altering QH, which cannot be changed in this scenario. The ideal is to approach the efficiency of a Carnot engine, which has the maximum possible efficiency between two given temperatures by operating in a reversible manner and reducing entropy generation.
A wooden artifact is found in an ancient tomb. Its 14C activity is measured to be 66.3% of that in a fresh sample of wood from the same region. Assuming the same amount of 14C was initially presented in the wood from which the artifact was made, determine the age of the artifact. The half-life of 14C is 5730 y. Answer in units of y.
Answer:
3396.53 years
Explanation:
Using decay formula
In([tex]\frac{N}{No}[/tex]) = -Kt where t is the age of the artifact in years and k is the decay constant
T1/2 = [tex]\frac{In2}{K}[/tex]
5730 = [tex]\frac{In2}{K}[/tex]
K = In 2 / 5730= 0.000121yr^-1
N / No = 0.663
In (0.663) / -0.000121 = t
t = 3396.53 years
Giraffe bending to drink. In a giraffe with its head 1.83 m above its heart, and its heart 2.04 m above its feet, the(hydrostatic) gauge pressure in the blood at its heart is 246 torr. Assume that the giraffe stands upright and the blood density is 1.06 × 103 kg/m3. In torr (or mm Hg), find the (gauge) blood pressure.
(a) at the brain (the pressure is enough to perfuse the brain with blood, to keep the giraffe from fainting)
(b) at the feet (the pressure must be countered by tight-fitting skin acting like a pressure stocking).
(c) If the giraffe were to lower its head to drink from a pond without splaying its legs and moving slowly, what would be the increase in the blood pressure in the brain? (Such action would probably be lethal.)
Answer:
1) Pm₂ = 1.9 10⁴ Pa , b) P_feet = 5.4 10⁴ Pa , c) Pm₄ = 4.4 10⁴ Pa
Explanation:
a) Pressure can be found using Bernoulli's equation
P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rgo g y₂
The amount of blood that runs through the constant system, all the blood that reaches the brain leaves it, so we can assume that the speed of entry and exit of the total blood is the same. In this case the equation is
P₁-P₂ = rgo h (y₂-y₁)
The gauge pressure is
Pm = P₁ -P₂
Pm₂ = 1.06 10³ 9.8 1.83
Pm₂ = 19 10³ Pa
Pm₂ = 1.9 10⁴ Pa
The pressure in the heart is
Pm₁ = 246 torr (1,013 10⁵ Pa / 760 torr) = 3,279 10⁴ Pa
Therefore the gauge pressure is an order of magnitude less
Total or absolute pressure is
Pm₂ = P_heart - P_brain
P_brain = P_heart - Pm₂
P brain = 3,279 10⁴ - 1.9 10⁴
P brain = 1.4 104 Pa
b) on the feet
Pm₃ = rho g y₃
y = 2.04 m
Pm₃ = 1.06 10³ 9.8 2.04
Pm₃ = 21 10³ Pa
Pm₃ = 2.1 10⁴ Pa
Total pressure
Pm₃ = P_feet + P_heart
P_feet = Pm₃ + P_heart
P_feet = 3,279 10⁴ + 2.1 10⁴
P_feet = 5.4 10⁴ Pa
c) If you lower your head the height change is
h = 1.83 +2.04
h = 4.23 m
Pm₄ = 1.06 10³ 9.8 4.23
Pm₄ = 4.4 10⁴ Pa
In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that the standing wave is produced by the superposition of traveling and reflected waves, where the incident traveling waves propagate in the +x direction with an amplitude A = 2.45 mm and a speed vx = 10.5 m/s . The first antinode of the standing wave is a distance of x = 27.5 cm from the left end of the string, while a light bead is placed a distance of 13.8 cm to the right of the first antinode. What is the maximum transverse speed vy of the bead? Make sure to use consistent distance units in your calculations.
Given that the distance from the left end of the string to the first antinode is 27.5 cm , calculate the wavelength of the standing wave on the string. Remember to convert all measurements into units of meters before performing this calculation.
Answer:
On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means
[tex]\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m[/tex]
This means that the relation between the wavelength and the length of the string is
[tex]3\lambda/2 = L[/tex]
By definition, this standing wave is at the third harmonic, n = 3.
Furthermore, the standing wave equation is as follows:
[tex]y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)[/tex]
The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.
[tex]v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)[/tex]
[tex]a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0[/tex]
For this equation to be equal to zero, sin(59.94t) = 0. So,
[tex]59.94t = \pi\\t = \pi/59.94 = 0.0524~s[/tex]
This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:
[tex]v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s[/tex]
In this exercise we have to use the knowledge of mechanics to be able to calculate the wavelength, so we can say that it will correspond to:
[tex]v_y=0.002m/s[/tex]
On the standing waves well-behaved, the first antinode happen one of four equal parts of a intuitiveness out completely. This means:
[tex]\lambda=1.1 m[/tex]
This way that the connection middle from two points the wavelength and the extent of object of the strand happen:
[tex]L=3\lambda /2[/tex]
By definition, this standing wave exist at the after second harmonious, n = 3.
Furthermore, the standing wave equating happen in this manner:
[tex]y(x,t)=(Asin(kx))sin(wt)=Asin(w/vx)sin(wt)=((2.45*10^{-3})sin(5.7X)sin(59.94t)[/tex]
The droplet exist established on x = 0.138 m. The maximum speed exist place the derivative of the speed function equals to nothing:
[tex]v_y=wAsin(kx)cos(wt)\\a_y=-w^2Asin(kx)sin(wt)\\a_y=0[/tex]
Find the time, we have:
[tex]59.94t=\pi\\t=0.0524s[/tex]
This exist moment of truth when the speed is maximum. So, the maximum speed maybe raise by stop up existing time into the velocity function:
[tex]v_y(x=0.138,t=0.0524)=(59.94)(2.45*10^{-3})sin((5.7)(0.138))cos((59.94)(0.0524))\\=0.002 m/s[/tex]
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In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg, roll toward each other and collide. The velocity is measured before and after each collision. The collected data is shown below. A 5 column table with 3 rows. The first column is unlabeled with entries Trial 1, Trial 2, Trial 3, Trial 4. The second column is labeled Initial Velocity Ball A (meters per second) with entries positive 1, positive 0.5, positive 2, positive 0.5. The third column is labeled Initial Velocity Ball B (meters per second) with entries negative 2, negative 1.5, positive 1, negative 1. The third column is labeled Final Velocity Ball A (meters per second) with entries negative 2, negative 0.5, positive 1, positive 1.5. The fourth column is labeled Final Velocity Ball B (meters per second) with entries negative 1, negative 0.5, negative 2, negative 1.5. Which trial shows the conservation of momentum in a closed system? Trial 1 Trial 2 Trial 3 Trial 4
Answer:
Second Trial satisfy principle of conservation of momentum
Explanation:
Given mass of ball A and ball B [tex]=\ 1.0\ Kg.[/tex]
Let mass of ball [tex]A[/tex] and [tex]B\ is\ m[/tex]
Final velocity of ball [tex]A\ is\ v_1[/tex]
Final velocity of ball [tex]B\ is\ v_2[/tex]
initial velocity of ball [tex]A\ is\ u_1[/tex]
Initial velocity of ball [tex]B\ is\ u_2[/tex]
Momentum after collision [tex]=mv_1+mv_2[/tex]
Momentum before collision [tex]= mu_1+mu_2[/tex]
Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.
Now, [tex]mu_1+mu_2=mv_1+mv_2[/tex]
Plugging each trial in this equation we get,
First Trial
[tex]mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3[/tex]
momentum before collision [tex]\neq[/tex] moment after collision
Second Trial
[tex]mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1[/tex]
moment before collision [tex]=[/tex] moment after collision
Third Trial
[tex]mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1[/tex]
momentum before collision [tex]\neq[/tex] moment after collision
Fourth Trial
[tex]mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0[/tex]
momentum before collision [tex]\neq[/tex] moment after collision
We can see only Trial- 2 shows the conservation of momentum in a closed system.
Answer: Trial 2
Explanation:
At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 750◦C and 440◦C respectively, and of the second 415◦C and 270◦C.
Answer:
The rate at which coal burned is 111.12 kg/s.
Explanation:
Given that,
First initial temperature =750°C
First final temperature =440°C
Second initial temperature =415°C
Second final temperature =270°C
Suppose If the heat of combustion of coal is [tex]2.8×10^{7}\ J/kg[/tex], at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65% of the Carnot efficiency.
The work done by first engine is
[tex]W=eQ[/tex]
The work done by second engine is
[tex]W'=e'Q'[/tex]
[tex]W'=e'Q(1-e)[/tex]
Total out put of the plant is given by
[tex]W+W'=950\ MW[/tex]
Put the value into the formula
[tex]eQ+e'Q(1-e)=950\times10^{6}[/tex]....(I)
We need to calculate the efficiency of first engine
Using formula of efficiency
[tex]e=65%(1-\dfrac{T_{L}}{T_{H}})[/tex]
[tex]e=0.65(1-\dfrac{440+273}{750+273})[/tex]
[tex]e=0.196[/tex]
We need to calculate the efficiency of second engine
Using formula of efficiency
[tex]e'=65%(1-\dfrac{T_{L}}{T_{H}})[/tex]
[tex]e'=0.65(1-\dfrac{270+273}{415+273})[/tex]
[tex]e'=0.136[/tex]
Put the value of efficiency for first and second engine in the equation (I)
[tex]Q(0.196+0.136(1-0.196))=950\times10^{6}[/tex]
[tex]Q=\dfrac{950\times10^{6}}{(0.196+0.136(1-0.196))}[/tex]
[tex]Q=3111.24\times10^{6}\ W[/tex]
We need to calculate the rate at which coal burned
Using formula of rate
[tex]R=\dfrac{Q}{H_{coal}}[/tex]
[tex]R=\dfrac{3111.24\times10^{6}}{2.8×10^{7}}[/tex]
[tex]R=111.12\ kg/s[/tex]
Hence, The rate at which coal burned is 111.12 kg/s.
Cold water (cp = 4180 J/kg·K) leading to a shower enters a thinwalled double-pipe counterflow heat exchanger at 15°C at a rate of 1.25 kg/s and is heated to 60°C by hot water (cp = 4190 J/kg·K) that enters at 100°C at a rate of 4 kg/s. If the overall heat transfer coefficient is 880 W/m2 ·K, determine the rate of heat transfer and the heat transfer surface area of the heat exchanger.
Answer:
the rate of heat transfer Q is Q =235.125 kJ/s
the heat transfer surface area A of the heat exchanger is A= 15.30 m²
Explanation:
Assuming negligible loss to the environment, then the heat flow of the hot water goes entirely to the cold water
Denoting a as cold water and b as hot water , then
Q= Fᵃ* cpᵃ * ( T₂ᵃ - T₁ᵃ)
where
F= mass flow
cp = specific heat capacity at constant pressure
T₂= final temperature
T₁ = initial temperature
replacing values
Q = Fᵃ* cᵃ * ( T₂ᵃ - T₁ᵃ) = 1.25 kg/s* 4180 J/kg·K* ( 60°C-15°C) * 1 kJ/1000J= 235.125 kJ/s
if all there is no loss to the surroundings
Qᵃ + Qᵇ = Q surroundings = 0
Fᵃ* cpᵃ * ( T₂ᵃ - T₁ᵃ) + Fᵇ* cpᵇ * ( T₂ᵇ - T₁ᵇ) = 0
T₂ᵇ = T₁ᵇ - [Fᵃ* cpᵃ / (Fᵇ* cpᵇ) ]* ( T₂ᵃ - T₁ᵃ)
replacing values
T₂ᵇ =100°C - [1.25 kg/s* 4180 J/kg·K/ (4 kg/s* 4190 J/kg·K)]* ( 60°C-15°C)
T₂ᵇ = 85.97 °C
the heat transfer surface of the heat exchanger is calculated through
Q = U*A* ΔTlm
where
U= overall heat transfer coefficient
A = heat transfer area of the heat exchanger
ΔTlm = (ΔTend - ΔTbeg)/ ln ( ΔTend - ΔTbeg)
ΔTbeg = temperature difference between the 2 streams at the inlet of the heat exchanger ( hot out - cold in) = 85.97 °C - 15°C = 70.97 °C
ΔTbeg = temperature difference between the 2 streams at the end of the heat exchanger ( hot in - cold out ) = 100°C - 60 °C = 40°C
then
ΔTlm = (ΔTend - ΔTbeg)/ ln ( ΔTend - ΔTbeg) =( 70.97 °C- 40°C)/ ln( 70.97°C/40°C) = 17.455 °C
ΔTlm = 17.455 °C
then
Q = U*A* ΔTlm
A = Q/(U*ΔTlm) = 235.125 kJ/s/(17.455 °C *880 W/m²*K) *1000 J/kJ = 15.30 m²
A= 15.30 m²
Final answer:
Calculate the heat transfer rate and heat transfer surface area in a double-pipe counterflow heat exchanger using given water properties and overall heat transfer coefficient.
Explanation:
Cold Water: mw = 1.25 kg/s, cp = 4180 J/kg·K, Tin = 15°C, Tout = 60°C
Hot Water: mh = 4 kg/s, cp = 4190 J/kg·K, Tin = 100°C, Tout = ?
Overall Heat Transfer: U = 880 W/m²·K
Calculate Heat Transfer Rate:
Calculate Q using Q = mcΔT for each water type.Calculate ∆T using Tin and Tout values.Use the overall heat transfer coefficient equation: Q = U × A × ∆Tlm.Solve for A, the heat transfer surface area.A ≈ 1m²
So, the rate of heat transfer is approximately 55341.28 W and the heat transfer surface area of the heat exchanger is approximately 1 m².
Since we cannot physically collect data from stars and most other objects in the universe, almost all of the information we obtain from the universe comes from analyzing the light, or spectra, from those objects. The study of light is known as spectroscopy. As we have seen in this simulation, every blackbody emits light with an easily identified pattern known as the blackbody curve. This is the particular way the total light emitted by a blackbody varies with its frequency. The exact form of the curve depends only on the body's temperature. Since we can treat stars as blackbodies, this is incredibly useful in astronomy that shows us that the color of a star is also indicative of its temperature. Use the simulation to determine the surface temperature of the following star: Betelgeuse is a red supergiant star in the constellation Orion. Knowing that Betelgeuse has peak intensity in the red and infrared wavelengths, adjust the intensity scale and temperature until you can determine the approximate surface temperature of the star. a. 3500K b. 4800K c. 7700K d. 11,000 K
Answer:
3500 K
Explanation:
b = Wien's displacement constant = [tex]2.89\times 10^{-3}\ mK[/tex]
Wavelength range = 700 nm to 10⁶ m. Let us take 825 nm
[tex]\lambda_m=825\ nm[/tex]
From Wien's displacement law we have
[tex]\lambda_m=\dfrac{b}{T}\\\Rightarrow T=\dfrac{b}{\lambda_m}\\\Rightarrow T=\dfrac{2.89\times 10^{-3}}{825\times 10^{-9}}\\\Rightarrow T=3500\ K[/tex]
The surface temperature of Betelguese is 3500 K
My Notes A container is divided into two equal compartments by a partition. One compartment is initially filled with helium at a temperature of 240 K; the other is filled with nitrogen at a temperature of 315 K. Both gases are at the same pressure. If we remove the partition and allow the gases to mix, what will be their final temperature?
Answer:
[tex]final-temperature = T_{f} = 252.51K[/tex]
Explanation:
we can solve this problem by using the first law of thermodynamics.
[tex]\Delta U= Q-W[/tex]
Q= heat added
U= internal energy
W= work done by system
[tex]E_{final}= E_{initial}[/tex]
[tex]C_{v} (N_{2}) C_{v}(He) T_{f} =C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}[/tex] (1)
[tex]C_{v}(N_{2})=1.04\frac{KJ}{Kg K}[/tex]
[tex]C_{v}(He)=5.193\frac{KJ}{Kg K}[/tex]
now
From equation 1
[tex]T_{f}=\frac{C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}}{C_{v} (N_{2}) C_{v}(He)}[/tex]
[tex]T_{f} = \frac{315\times1.04+5.193\times240}{1.04+5.193}[/tex]
[tex]T_{f} = 252.51K[/tex]
Select the statement that correctly completes the description of phase difference.
Phase difference describes:
O the difference in the phase angle between any two waves at any given position along the waves.
O the shift between the positions of corresponding crests of two waves of the same frequency.
O the difference in the frequencies of two waves at a given time.
O the displacement of a wave particle from its undisturbed position at the origin.
Phase difference denotes the difference in phase angle between two waves at a given point, occurring when waves are separated by a whole number of multiples of wavelengths.
Explanation:Phase difference describes the difference in the phase angle between any two waves at any given position along the waves. When the waves have the same frequency and the difference in their path lengths is an integer multiple of the wavelength, the waves are said to be in phase. This means these points are separated by a whole number multiple of whole wave cycles or wavelengths. For example, sound waves can illustrate a phase shift when they have different path lengths. It is also important to understand that the wavelength is defined as the distance between any two adjacent points that are in phase.
Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheels energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg.
1. A motor spins up the flywheel with a constant torque of 50 N*m. How long does it take the flywheel to reach top angular speed of 1200 rpm
2. How much energy is stored in the flywheel?
Final answer:
To find the time it takes for the flywheel to reach its top angular speed, we can use the formula: ω = Δθ / Δt. In this case, the flywheel has a diameter of 1.5 m and spins up to an angular speed of 1200 rpm. It takes approximately 0.0375 seconds for the flywheel to reach its top angular speed. The energy stored in the flywheel is approximately 554,414.06 joules.
Explanation:
To find the time it takes for the flywheel to reach its top angular speed, we can use the formula:
ω = Δθ / Δt
Where ω is the angular velocity, Δθ is the change in angle, and Δt is the change in time.
In this case, the flywheel has a diameter of 1.5 m, so the radius is 0.75 m. The flywheel spins up from rest to an angular speed of 1200 rpm, which is equivalent to 125.66 rad/s.
Using the formula, we can rearrange to solve for Δt:
Δt = Δθ / ω
Δθ is equal to the circumference of the flywheel, which is 2π times the radius:
Δθ = 2π × 0.75 m = 4.7124 rad
Plugging in the values:
Δt = 4.7124 rad / 125.66 rad/s = 0.0375 s
So it takes approximately 0.0375 seconds for the flywheel to reach its top angular speed.
Now, to calculate the energy stored in the flywheel, we can use the formula:
KE = 0.5 × I × ω^2
Where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia for a solid disk is given by:
I = 0.5 × m × r^2
Where m is the mass of the flywheel and r is the radius.
Plugging in the values:
I = 0.5 × 250 kg × (0.75 m)^2 = 70.3125 kg×m^2
Now we can calculate the energy:
KE = 0.5 × 70.3125 kg×m^2 × (125.66 rad/s)^2 = 554,414.06 J
So, the flywheel stores approximately 554,414.06 joules of energy.
You are in a submarine and are at the surface of the ocean but out in the deep sea. There is a big storm and you want to dive down deep enough so that you avoid the turbulence of the sea. You notice the distance between successive waves (wave length) is 20 meters. How far down do you need to dive down to not feel the effect of the waves?
One of the maritime principles that relate the turbulence and wavelength of the waves is called the "depth of 1/2 wavelength" which is also usually referred to as the floor of the wave: A point of depth in which There is no movement. There if a submarine is found, it can be unbalanced and steadily navigate.
If the wavelength is 20 meters, then it must be submerged 10 meters (20/2) to avoid turbulence.
A 44.5 mA current is carried by a uniformly wound air-core solenoid with 500 turns, a 18.5 mm diameter, and 14.0 cm length. (a) Compute the magnetic field inside the solenoid. µT (b) Compute the magnetic flux through each turn. T·m2 (c) Compute the inductance of the solenoid. mH (d) Which of these quantities depends on the current? (Select all that apply.)
Answer:
a. Magnetic Field =1.997×[tex]10^{-4}[/tex] T
b. Area= 2.68×[tex]10^{-4}[/tex][tex]m^{2}[/tex]
Magnetic Flux= 5.367×[tex]10^{-8}[/tex]T[tex]m^{2}[/tex]
c. Inductance= 6.013×[tex]10^{-4}[/tex]H
Explanation:
Parameters from the question
I= 44.5×[tex]10^{-3}[/tex]A
N=500 turns
Diameter=18.5mm
Radius = (diameter/2) = 9.25mm =9.25×[tex]10^{-3}[/tex]m
L= 14cm = 0.14m
Permitivity [tex]U_{o}[/tex]=4π×[tex]10^{-7}[/tex]H/m
The Formulars Used are
B(Magnetic Field) =[tex]\frac{U_{o}. N. I }{l}[/tex]
Mag Flux= B.A
Inductance= [tex]\frac{U_{o}.N^{2} .A }{l}[/tex]
Answer:
a) 199.716 μT
b) [tex]5.368 * 10^{-8}[/tex] T·m^2
c) 0.603 mH
d) B and Ф
Explanation:
I am giving the explanation with my handwritten solution in the paper.
Check the attachment please.
A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2.4 cm thick on the inside wall surface. The wood has k=0.080W/(m?K), and the Styrofoam has k= 0.010 W/(m?K). The interior surface temperature is 20.0 ?C , and the exterior surface temperature is -13.0 ?C
A.)What is the temperature at the plane where the wood meets the Styrofoam? _______ Celsius
B.)What is the rate of heat flow per square meter through this wall? ______W/m^2
Answer:
Explanation:
Given
thermal conductivity of wood [tex] K_w=0.08 W/m^2-K[/tex]
thermal conductivity of insulation [tex]K_i=0.01 W/m^2-K[/tex]
thickness of wood [tex]t_2=3 cm[/tex]
thickness of insulation [tex]t_1=2.4 cm[/tex]
[tex]T_i=20^{\circ}C[/tex]
[tex]T_o=-13^{\circ}C[/tex]
we know heat Flow is given by
[tex]Q=kA\frac{dT}{dx}[/tex]
[tex]dT=[/tex] change in temperature
[tex]dx=[/tex] thickness
K=thermal conductivity
A=Area of cross-section
A is same
Suppose T is the temperature of Junction
as heat Flow is same thus
[tex]\frac{k_w(20-T)}{3}=\frac{k_i(T-(-13))}{2.4} [/tex]
[tex]\frac{0.08(20-T)}{3}=\frac{0.01(T+13)}{2.4}[/tex]
[tex]T=19.36 ^{\circ}C[/tex]
(b)Rate of heat flow
[tex]Q=\frac{k_w(T+13)}{3\times 10^{-2}}[/tex]
[tex]Q=\frac{0.08\times 32.36}{0.03}[/tex]
[tex]Q=86.303 W/m^2[/tex]
The temperature at which the wood meets the Styrofoam is determined using the proportional temperature drops across each material, considering their respective thermal resistances. The rate of heat flow per square meter through the wall is calculated using Fourier's law, considering the combined thermal resistance of both layers.
Explanation:Temperature at the Wood-Styrofoam Plane and Heat Flow Rate
To find the temperature at the plane where the wood meets the Styrofoam, we need to use the concept of thermal resistance and the fact that the heat flow through both materials is the same. With the given thermal conductivity (k values) and thicknesses of the wood and Styrofoam, we can calculate their respective thermal resistances (R = thickness/k). Then, by setting up a proportion, we can find the temperature drop across the wood (ΔTw) and the Styrofoam (ΔTs) using the formula ΔT = (k*A* ΔT)/(thickness), where A is the area (which cancels out as it's the same for both layers). Knowing the temperature drop across each and the outer and inner surface temperatures, we can find the temperature at the plane where the layers meet.
To calculate the rate of heat flow per square meter through this wall, we'll use Fourier's law of thermal conduction, which is given by Q = k*A* ΔT/d, where Q is the heat flow rate, A is the area, ΔT is the temperature difference, and d is the thickness. Since the wall consists of two layers with different thermal conductivities and thicknesses, we need to calculate the equivalent thermal resistance for the combined wall and then use the overall temperature difference to find the heat flow rate.
An ideal heat engine operates of fixed difference of temperatures between hot and cold reservoirs, say 100K. What will provide greater efficiency: operation with the hot reservoir as hot as possible, or operation with a cold reservoir as cold as possible?
To solve this problem we will apply the concepts given for the efficiency of an engine which is given as
[tex]\eta = 1-\frac{T_C}{T_H}[/tex]
[tex]\eta = \frac{T_H-T_C}{T_H}[/tex]
Where
[tex]T_C[/tex] = Temperature of the cold reservoir
[tex]T_H[/tex] = Temperature of the hot reservoir
The efficiency maximum would be given only if [tex]T_C = 0[/tex]
Replacing this value we have
[tex]\eta = \frac{T_H-0}{T_H}[/tex]
[tex]\eta = 1[/tex]
Therefore: Cold reservoir as cold as possible provide the greater efficiency.
Final answer:
The greater efficiency in an ideal heat engine is obtained when the hot reservoir is as hot as possible and the cold reservoir is as cold as possible.
Explanation:
The greater efficiency in an ideal heat engine operating with a fixed temperature difference between hot and cold reservoirs is obtained when the hot reservoir is as hot as possible and the cold reservoir is as cold as possible.
Efficiency is determined by the ratio of the temperature of the cold reservoir (Tc) to the temperature of the hot reservoir (Th). The greater the temperature difference, the easier it is to convert heat transfer to work.
Therefore, maximizing the temperature difference by making the hot reservoir as hot as possible and the cold reservoir as cold as possible will result in greater efficiency.
A grinding wheel is a uniform cylinder with a radius of 7.80 cm and a mass of 0.550 kg.
Part A
Calculate its moment of inertia about its center. Express your answer to three significant figures and include the appropriate units.
Part B
Calculate the applied torque needed to accelerate it from rest to 1750 rpm in 7.40 s .
Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 58.0 s .
Answer:
a. I = 167.31 x 10 ⁻³ kg*m²
b. T = 4.59 kg * m² / s²
Explanation:
The moment of inertia of a uniform cylinder:
a.
r = 7.8 cm * 1 m / 100 cm = 0.078 m
I = ½ * m * r²
I = ½ * 0.55 kg * (0.078²m)
I = 167.31 x 10 ⁻³ kg*m²
b.
T = Iα’ + Iα,
α’ = ω’/t = 1750 rpm * (2π/60) / 7.40s = 24.76 rad/s²
α = ω/t = 1500 rpm * (2π/60) / 58 = 2.71 rad/s²
T = (167.31 x 10⁻³ kg*m²)* (24.76 + 2.71 ) rad / s²
T = 4.59 kg * m² / s²
The moment of inertia of the wheel is calculated as 0.00133 kg*m^2. The second part of the question involves determining the net and frictional torques to find the total applied torque.
Explanation:To solve this problem, we need to apply the formulas of moment of inertia and the angular acceleration along with the concept of frictional torque. The moment of inertia for a cylinder rotating about its axis is given by the formula I = 0.5*m*r^2. In this case, where mass (m) is 0.550 kg and radius (r) is 7.80 cm or 0.078 m (since 1cm = 0.01m).
Part A: I = 0.5 * 0.550 kg * (0.078 m)^2 = 0.00133 kg*m^2.
For Part B, we first need to convert the rotational speed from revolutions per minute (rpm) to rad/s. Then we use these values to determine the angular acceleration and calculate the net torque. The frictional torque is then added to this net torque to find the total applied torque.
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A 92kg astronaut and a 1200kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving it a speed of 0.14m/s directly away from the shuttle. Seven and a half seconds later the astronaut comes into contact with the shuttle. What was the initial distance from the shuttle to the astronaut?
Answer:
13.7m
Explanation:
Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.
After the push
[tex]m_av_a + m_sv_s = 0[/tex]
Where [tex]m_a = 92kg[/tex] is the mass of the astronaut, [tex]m_s = 1200kg[/tex] is the mass of the satellite, [tex]v_s = 0.14 m/s[/tex] is the speed of the satellite. We can calculate the speed [tex]v_a[/tex] of the astronaut:
[tex]v_a = \frac{-m_sv_s}{m_a} = \frac{-1200*0.14}{92} = -1.83 m/s[/tex]
So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be
d = vt = 1.83 * 7.5 = 13.7 m
This is a conservation of momentum problem where the astronaut moves opposite to the direction of the satellite's movement due to Newton's third law. The astronaut's velocity is calculated using the conservation of momentum principle, and the distance between him and the shuttle is then determined via the formula for distance.
Explanation:This problem involves understanding the conservation of momentum in a system with no external forces acting on it. When an astronaut pushes a satellite in space, there's a reaction force acting back on the astronaut due to Newton's third law. So, the astronaut will also move in the opposite direction. Keep in mind that the net momentum before and after this action remains zero as there are no external forces.
We're given that the astronaut comes into contact with the shuttle seven and half seconds after pushing on the satellite. He must have been moving at a certain speed to cover the distance in this time. Due to conservation of momentum, we can set up an equation as follows: Momentum of Astronaut + Momentum of Satellite = 0 (Because initially they were at rest). We can then calculate this to find the velocity of the astronaut.
After getting the velocity of the astronaut, we use the formula for distance: Distance = Speed * Time to get the initial distance between astronaut and Shuttle.
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Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K. You set the bowl up so that as it cools to room temperature the heat first flows through a Carnot Engine. The soup has Cv= (33 J/K). Assume that the volume of the soup does not change.
1. What fraction of the total heat QH that is lost by the soup can be turned into useable work by the engine?
Answer:
Explanation:
Heat energy given out by the soup
= C_v x ( t₂ - t₁ )
= 33 x ( 340 - 300)
= 1320 J
This heat is given to Carnot engine . Efficiency of engine
= (340 - 300 ) / 340
= 40 / 340
2 / 17
This fraction of total heat given is converted into useable work by the engine.
The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours (A?h). A 50 A?h battery can supply a current of 50 Afor 1.0 h, or 25 A for 2.0 h, and so on.A) What total energy can be supplied by a 13V , 60A?h battery if its internal resistance is negligible?Answer= ...... JB) What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (See Section 17.6; the density of gasoline is 90 kg/m 3.)Answer= ........ LC) If a generator with an average electrical power output of 0.45 kW is connected to the battery, how much time will be required for it to charge the battery fully?Answer= ........ h
Answer: (A) 780J
(B) 1.89×10^-11L
(C)1.67×10^-4 h
Explanation:
Energy of the battery = IVt
=13×60 = 780J
Heat combustion of
1g of gasoline relax 46000J
Therefore 780J will release 780/46000
= 0.017g
Density = mass/volume
Volume = mass/density
Volume =0.017× 10^-3 / 900
= 1.89× 10^-8 m3
= 1.89×10^-11 litres
P=IVt
t=P/IV
= 450/60×13
1.67×10^-4 hours
The energy of a battery, volume of gasoline and time required to charge the battery is required.
The energy is 2808000 J
The volume is 0.683 L.
The time required is 1.733 h
It = Current-time = 60 Ah = [tex]60\times 3600\ \text{As}[/tex]
t = Time
V = Voltage = 13 V
Energy is given by
[tex]E=IVt=ItV\\\Rightarrow E=60\times 3600\times 13\\\Rightarrow E=2808000\ \text{J}[/tex]
[tex]\rho[/tex] = Density = [tex]90\ \text{kg/m}^3=\dfrac{90}{1000}=0.09\ \text{kg/L}[/tex]
C = Thermal heat capacity = [tex]4.57\times 10^7\ \text{J/kg}[/tex]
m = Mass
Power is given by
[tex]P=mC\\\Rightarrow m=\dfrac{P}{C}\\\Rightarrow m=\dfrac{2808000}{4.57\times 10^7}=\dfrac{702}{11425}\ \text{kg}[/tex]
Volume is given by
[tex]V=\dfrac{m}{\rho}\\\Rightarrow V=\dfrac{\dfrac{702}{11425}}{0.09}\\\Rightarrow V=0.683\ \text{L}[/tex]
P = Power = 0.45 kW
Time is given by
[tex]t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{2808000}{450}=6240\ \text{s}\\\Rightarrow t=\dfrac{6240}{3600}=1.733\ \text{h}[/tex]
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The SI unit of power is the watt. Which of the following units are equivalent to the watt?
A) V∙AB) J/CC) C/sD) V/sE) A/s
Answer:
The right option is (A) V.A
Explanation:
Power: This is the rate at which work is done. Or it is the produce of force and velocity. The S.I unit of power is Watt (W). Other units include Horse power(hp), foot-pound per minutes, etc.
Generally, power can be represented as,
Power = Energy/time
P = W/t......................... Equation 1
Where p = power, w = Work or energy, t = time in seconds.
Electrical energy: This is the product of potential difference and the quantity of charge.
∴ W = VQ............................... Equation 2
Where V = potential difference, Q = quantity of charge and W = Energy or Work done.
Also Q = It........................ Equation 3.
where I = current in ampere, t = time in seconds
Substituting equation 3 into equation 3
W = VIt............................ Equation 4.
Also substituting Equation 4 into Equation 1
P = VIt/t = VI = voltage(V)×Current(A)
Therefore the equivalent unit of power is
P = V.A.
The right option is (A) V.A
The SI unit of power is the watt. The options A) V∙A, B) J/C, and C) C/s are equivalent to a watt.
Explanation:The SI unit of power is indeed the watt, represented by the symbol 'W'. The watt is a derived unit of power in the International System of Units (SI) and is defined as one joule per second. Hence, three of the given options, A) V∙A, B) J/C, and C) C/s are equivalent to a watt.
A) A Volt (V) times an Ampere (A) also equals a watt (V∙A=W). This is derived from the formula P=V∙I where P is power, V is voltage, and I is current. B) A Joule (J) per Coulomb (C) is also a watt (J/C = W). This comes from the relationship P=W/t = J/s = V∙A. C) A Coulomb (C) per second is also a watt (C/s = W), because one ampere equals to 1 C/s. The unit Coulomb per second refers to the electrical current where 1 A is equivalent to 1 C/s.Learn more about Watt here:https://brainly.com/question/27355276
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The period of a sinusoidal source is the time required for the sinusoid to pass through all of its possible values. We use the symbol T to represent the period of a sinusoid. The period and the frequency are inversely related. A sinusoidal source described by the function cos(ωt) has a frequency of ω radians/second, or a frequency f=ω/2π Hz. The units hertz represents the number of cycles per second. Since the period is the number of seconds per cycle, the period is the inverse of the frequency in hertz: T=1f Substituting the frequency in radians/second, ω, for the frequency in Hz gives us another way to calculate the period: T=2πω What is the period of the voltage source described as v(t)=50cos(2000t−45∘) mV? Express your answer to two digits after the decimal point and include the appropriate units.
Answer:
T=0.0031secs
Explanation:
The voltage expression [tex]v(t)=50cos(2000t-45^{0})[/tex] can be represented as
[tex]v(t)=v_{m}cos(wt-\alpha ) \\[/tex]
comparing the two equations we can conclude that the angular frequency
[tex]w=2000[/tex]
from the question, since the frequency,f which is express as
[tex]f=\frac{w}{2\pi }\\[/tex],
Hence [tex]f=\frac{2000}{2\pi } \\f=\frac{2000}{2*3.14 } \\f=318.471Hz\\[/tex].
The period which is the inverse of the frequency can be express as
[tex]T=\frac{1}{f} \\T=\frac{1}{314.471}\\ T=0.00314\\T=0.0031secs[/tex]