Answer:
Explanation:
Given
mass of bullet [tex]m=7 gm[/tex]
mass of gun and Puck [tex]M=150 gm[/tex]
speed of gun and Puck is [tex]v=0.53 m/s[/tex]
Let speed of bullet be u
conserving Momentum
initial momentum=Final Momentum
[tex]0=Mv+mu[/tex]
[tex]u=-\frac{M}{m}v[/tex]
[tex]u=-\frac{150}{7}\times 0.53=-11.35 m/s[/tex]
negative sign indicates that bullet is moving in opposite direction
Using the law of conservation of momentum, it's possible to calculate that the bullet moves in the opposite direction of the gun and puck at a speed of approximately 113.6 m/s.
Explanation:This question represents a practical example of the conservation of momentum. According to the law of conservation of momentum, the total momentum of an isolated system remains constant if no external forces act on it. As the system (gun and bullet) is initially at rest, the total momentum is zero, and it remains zero even after the bullet is fired.
Let's denote the bullet's velocity as v. The momentum of the bullet after firing is its mass times its velocity, or 0.007kg * v. The momentum of the gun and puck is their combined mass times their velocity, or 0.150kg * 0.530m/s. As the total momentum should remain zero, these two quantities must be equal and opposite. Thus, 0.007kg * v = - 0.150kg * 0.530m/s. Solving for v, we find that the bullet's speed is approximately -113.6 m/s. The negative sign indicates that the bullet moves in the opposite direction of the gun and puck.
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A rigid, insulated tank that is initially evacuated is connected though a valve to a supply line that carries steam at 1 MPa and 300∘C. Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa, at which point the valve is closed.
Determine the final temperature of the steam in the tank, in ∘C.
Answer:
Final Temperature of the steam tank = 456.4°C
Explanation:
Assuming it to be a uniform flow process, kinetic and potential energy to be zero, and work done and heat input to be zero also. We can conclude that,
Enthalpy of the steam in pipe = Internal Energy of the steam in tank
Using the Property tables and Charts - Steam tables,
At Pressure= 1 MPa and Temperature= 300°C,
Enthalpy = 3051.2 kJ/kg
At Pressure= 1 MPa and Internal Energy= 3051.2 kJ/kg,
Temperature = 456.4°C.
A 1.53-kg piece of iron is hung by a vertical ideal spring. When perturbed slightly, the system is moves up and down in simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm. If we choose the total potential energy (elastic and gravitational) to be zero at the equilibrium position of the hanging iron, what is the total mechanical energy of the system?
A. 0.844 JB. 0.646 JC. 0.000 JD. 0.955 JE. 0.633 J
To solve this problem it is necessary to apply the concepts related to the elastic potential energy from the simple harmonic movement.
Said mechanical energy can be expressed as
[tex]E = \frac{1}{2} kA^2[/tex]
Where,
k = Spring Constant
A = Cross-sectional Area
From the angular movement we can relate the angular velocity as a function of the spring constant and the mass in order to find this variable:
[tex]\omega^2 = \frac{k}{m}[/tex]
[tex]k = m\omega^2\rightarrow \omega = 2\pi f [/tex] for f equal to the frequency.
[tex]k = 1.53(2\pi 1.95)^2[/tex]
[tex]k = 229.44[/tex]
Finally the energy released would be
[tex]E = \frac{1}{2} (229.44)(0.075)^2[/tex]
[tex]E = 0.6453J \approx 0.646J[/tex]
Therefore the correct answer is B.
The correct answer is option A. The total mechanical energy of the system is [tex]\( 0.844 \, \text{J} \)[/tex]
To find the total mechanical energy of the system, we need to calculate both the elastic potential energy (due to the spring) and the gravitational potential energy (due to the height above the equilibrium position) at the maximum amplitude point.
First, let's calculate the spring constant [tex]\( k \)[/tex] using the formula for the frequency[tex]\( k \)[/tex] of a mass-spring system:
[tex]\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \][/tex]
where \( m \) is the mass of the iron piece. Solving for [tex]\( k \)[/tex], we get:
[tex]\[ k = (2\pi f)^2 m \][/tex]
[tex]\[ k = (2\pi \times 1.95 \, \text{Hz})^2 \times 1.53 \, \text{kg} \][/tex]
[tex]\[ k \approx 244.73 \, \text{N/m} \][/tex]
Next, we calculate the elastic potential energy [tex]\( U_{\text{elastic}} \)[/tex] at the maximum amplitude [tex]\( A \)[/tex]:
[tex]\[ U_{\text{elastic}} = \frac{1}{2} k A^2 \][/tex]
[tex]\[ U_{\text{elastic}} = \frac{1}{2} \times 244.73 \[/tex], [tex]\text{N/m} \times (0.0750 \[/tex], [tex]\text{m})^2 \] \[ U_{\text{elastic}} \approx 0.683 \[/tex], [tex]\text{J} \][/tex]
Now, let's calculate the gravitational potential energy [tex]\( U_{\text{gravitational}} \)[/tex]at the maximum amplitude:
[tex]\[ U_{\text{gravitational}} = m g h \][/tex]
where [tex]\( g \)[/tex]is the acceleration due to gravity (approximately [tex]\( 9.81 \, \text{m/s}^2 \[/tex])) and [tex]\( h \)[/tex] is the height at maximum amplitude. Since the amplitude is given in centimeters, we convert it to meters:
[tex]\[ h = 7.50 \, \text{cm} = 0.0750 \, \text{m} \][/tex]
[tex]\[ U_{\text{gravitational}} = 1.53 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.0750 \, \text{m} \][/tex]
[tex]\[ U_{\text{gravitational}} \approx 1.130 \, \text{J} \][/tex]
The total potential energy [tex]\( U_{\text{total}} \)[/tex] at the maximum amplitude is the sum of the elastic and gravitational potential energies:
[tex]\[ U_{\text{total}} = U_{\text{elastic}} + U_{\text{gravitational}} \] \[ U_{\text{total}} \approx 0.683 \, \text{J} + 1.130 \, \text{J} \] \[ U_{\text{total}} \approx 1.813 \, \text{J} \][/tex]
However, we must consider that the total potential energy is zero at the equilibrium position. Therefore, the total mechanical energy [tex]\( E \)[/tex] of the system is equal to the total potential energy at the maximum amplitude:
[tex]\[ E = U_{\text{total}} \] \[ E \approx 1.813 \, \text{J} \][/tex]
Since the potential energy at the equilibrium position is zero, the total mechanical energy is simply the sum of the elastic and gravitational potential energies at the maximum amplitude, which is approximately [tex]\( 1.813 \, \text{J} \)[/tex]. However, the options given are in the range of [tex]\( 0.633 \, \text{J} \) to \( 0.955 \, \text{J} \)[/tex], which suggests that there might be a mistake in the calculation or interpretation of the problem.
Upon re-evaluating the calculation, it seems that the gravitational potential energy was incorrectly calculated. The correct calculation should consider that the gravitational potential energy is zero at the equilibrium position, not at the maximum amplitude. Therefore, we should only consider the change in gravitational potential energy from the equilibrium to the maximum amplitude, which is half the amplitude squared divided by the spring constant, as the mass will oscillate symmetrically around the equilibrium position.
The correct gravitational potential energy [tex]\( U_{\text{gravitational}} \)[/tex] at the maximum amplitude is:
[tex]\[ U_{\text{gravitational}} = \frac{1}{2} k \left(\frac{h}{2}\right)^2 \][/tex]
[tex]\[ U_{\text{gravitational}} = \frac{1}{2} \times 244.73 \, \text{N/m} \times \left(\frac{0.0750 \, \text{m}}{2}\right)^2 \][/tex]
[tex]\[ U_{\text{gravitational}} \approx 0.160 \, \text{J} \][/tex]
Now, the total mechanical energy [tex]\( E \)[/tex] of the system is:
[tex]\[ E = U_{\text{elastic}} + U_{\text{gravitational}} \] \[ E \approx 0.683 \, \text{J} + 0.160 \, \text{J} \] \[ E \approx 0.844 \, \text{J} \][/tex]
A certain transverse wave is described by y(x,t)=Bcos[2π(xL−tτ)],where B = 6.40 mm , L = 26.0 cm , and τ = 3.90×10^−2 s .a. Determine the wave's amplitude.b. Determine the wave's wavelength.c. Determine the wave's frequency.d. Determine the wave's speed of propagation. e. Determine the wave's direction of propagation.
Answer
given,
[tex]y(x,t)=B cos[2\pi (\dfrac{x}{L} - \dfrac{t}{\tau})][/tex]
B = 6.40 mm , L = 26 cm , τ = 3.90 × 10⁻² s
general wave equation
y = A cos (k x - ωt)
where A is the amplitude of the
a) Amplitude of the given wave
B = 6.40 mm
b) Wavelength of the given wave
λ = L
λ = 26 cm
c) wave frequency
[tex]f = \dfrac{1}{\tau}[/tex]
[tex]f = \dfrac{1}{3.9 \times 10^{-2}}[/tex]
f = 25.64 Hz
d) speed of wave will be equal to
v = f λ
v = 25.64 x 0.26
v = 6.67 m/s
e) direction of propagation will be along +ve x direction because sign of k x and ωt is same as general equation.
Because sunspots are a little cooler than the average temperature of the photosphere, they prevent some energy from being released from the part of the surface they occupy, but this energy is usually released from hotter and brighter than average areas nearby. Please answer the question: What could happen to the Sun if this energy release did not happen
Answer: Corona Mass Ejections(CME) and Solar flares
Explanation: Corona Mass Ejections and Solar flares are eruptions that occur in the sun due to the instability in the magnetic field of the sun. This Corona Mass Ejections and Solar flares are prevented by sun spots.Corona mas Ejections are Large and massive eruptions,solar flares are somewhat small eruptions,both of them are prevented from occuring by Sunspots which help to dissipate cooler temperatures in the sun.
Cart 111 of mass mmm is traveling with speed v_0v 0 v, start subscript, 0, end subscript in the + x+xplus, x-direction when it has an elastic collision with cart 222 of mass 3m3m3, m that is at rest. What are the velocities of the carts after the collision?
Answer:
In an elastic collision, the momentum and the kinetic energies are conserved.
Momentum:
[tex]\vec{P_i} = \vec{P_f}\\\vec{P}_1 + \vec{P}_2 = \vec{P}_1' + \vec{P}_2'\\m\vec{v_0} + 0 = m\vec{v_1}' + 3m\vec{v_2}}' \\v_0 = v_1 + 3v_2[/tex]
Kinetic energy:
[tex]K_i = K_f\\K_1 + K_2 = K_1' + K_2'\\\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}m{v_1'}^2 + \frac{1}{2}3m{v_2'}^2\\v_0^2 = {v_1'}^2 + 3{v_2'}^2[/tex]
We have two equations and two unknowns:
[tex]v_0 = v_1' + 3v_2'\\v_0^2 = {v_1'}^2 + 3{v_2'}^2\\\\3v_2' = v_0 - v_1'\\3{v_2'}^2 = {v_0}^2 - {v_1'}^2\\\\3{v_2'}^2 = (v_0 - v_1')(v_0 + v_1') = 3{v_2}'(v_1' + v_0)\\\\v_2' = v_1' + v_0\\3v_2' = v_0 - v_1'\\\\4v_2' = 2v_0\\\\v_2' = v_0/2\\v_1' = -v_0/2[/tex]
Explanation:
The first cart hits the second cart at rest and turns back with half its speed.
The second cart starts moving to the right with half the initial speed of the first cart.
Answer:
v1 = -v0/3 ,v2 = 2v0/3
Explanation:
A machine gear consists of 0.10 kg of iron and 0.16 kg of copper.
How much total heat is generated in the part if its temperature increases by 35 C°? (Specific heats of iron and copper are 450 and 390 J/kg⋅°C, respectively.)
a. 910 Jb. 3 800 Jc. 4 000 Jd. 4 400 J
Answer:
option (c)
Explanation:
mass of iron = 0.10 kg
mass of copper = 0.16 kg
rise in temperature, ΔT = 35°C
specific heat of iron = 450 J/kg°C
specific heat of copper = 390 J/kg°C
Heat by iron (H1) = mass of iron x specific heat of iron x ΔT
H1 = 0.10 x 450 x 35 = 1575 J
Heat by copper (H2) = mass of copper x specific heat of copper x ΔT
H1 = 0.16 x 390 x 35 = 2184 J
Total heat H = H1 + H2
H = 1575 + 2184 = 3759 J
by rounding off
H = 4000 J
Answer:b
Explanation:
Given
mass of iron Gear [tex]m_{iron}=0.1 kg[/tex]
mass of copper [tex]m_{cu}=0.16 kg[/tex]
specific heat of iron [tex]c_{iron}=450 J/kg-^{\circ}C[/tex]
specific heat of copper [tex]c_{cu}=390 J/kg-^{\circ}C[/tex]
heat gain by iron gear [tex]=m_{iron}c_{iron}\Delta T[/tex]
[tex]Q_1=0.1\times 450\times 35=1575 J[/tex]
heat gain by iron gear [tex]=m_{cu}c_{cu}\Delta T[/tex]
[tex]Q_2=0.16\times 390\times 35=2184 J[/tex]
Total heat [tex]Q_{net}=Q_1+Q_2[/tex]
[tex]Q_{net}=1575+2184=3759 J\approx 3800 J[/tex]
Your heart pumps blood at a pressure of 100 mmHg and flow speed of 60 cm/s. At your brain, the blood enters capillaries with such a large total cross-sectional area that the blood velocity is much smaller: about 0.7 mm/s. Ignore viscosity in the capillaries. How long could your neck be so that blood still reaches your brain in m
The length of the neck that blood can still reach the brain depends on blood pressure, flow speed, and the cross-sectional area of capillaries. Using the given values, the approximate length of the neck is 9.8 meters.
Explanation:The length of your neck that blood can still reach your brain depends on the blood pressure, flow speed, and the cross-sectional area of the capillaries. The blood enters the capillaries in the brain with a much smaller velocity of about 0.7 mm/s due to the large total cross-sectional area. To determine the length of the neck, we need to consider the time it takes for blood to travel from the heart to the brain. Using the given velocity, we can calculate that the time it takes for blood to reach the brain is approximately 14 seconds. Assuming a constant velocity, the length of the neck can be calculated using the formula: length = velocity x time, which gives us a result of 9.8 meters.
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10 kg of liquid water is in a container maintained at atmospheric pressure, 101325 Pa. The water is initially at 373.15 K, the boiling point at that pressure.The latent heat of water -> water vapor is 2230 J/g. The molecular weight of water is 18 g.103 J of heat is added to the water.1)How much of the water turns to vapor?mass(vapor)=
Answer:
[tex]m=0.0462\ g[/tex] of water is converted into vapour.
Explanation:
Given:
mass of water, [tex]m_w=10\ kg[/tex]pressure conditions, [tex]P=101325\ Pa[/tex]temperature conditions, [tex]T=373.15\ K[/tex]latent heat of vapourization of water, [tex]L=2230\ J.g^{-1}[/tex]amount of heat supplied to the water, [tex]103\ J[/tex]Now using the equation of heat considering latent heat only:
(since water already at boiling point at atmospheric temperature)
[tex]Q=m.L[/tex]
[tex]103=m\times 2230[/tex]
[tex]m=0.0462\ g[/tex] of water is converted into vapour.
In the situation provided, about 46.2 grams of the water will have transitioned from a liquid to a vapor after being supplied with 103 kJ of energy, given the specified latent heat of vaporization.
Explanation:Given that the latent heat of water's vaporization is 2230 J/g and 103 J of energy was provided to the water, we first convert all our units to be consistent. Remember that the latent heat of vaporization is the amount of heat energy required to change one gram of a substetance from a liquid to a gas at constant mperature and pressure. In this case, we're transitioning water to water vapor.
The input energy is 103 kJ, and the latent heat of vaporization is 2.23 kJ/g, so we can calculate the mass of the water that was vaporized using the equation: mass (g) = energy input (kJ) / latent heat of vaporization (kJ/g). By plugging in the values we get: mass = 103 / 2.23 = 46.2 grams.
So, approximately 46.2 grams of the water will have transitioned from a liquid to a vapor given the provided energy input of 103 kJ.
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In the fastest measured tennis serve, the ball left the racquet at 73.14 m/s. A serve tennis ball is typically in contact with the racquet for 30.0 ms and starts from rest. Assume constant acceleration.(a) what was the ball's acceleration during this serve??(b) how far did the ball travel during the serve???
Answer:
a) the acceleration is a= 2438 m/s²
b) the distance travelled during serve is d = 1.0971 m
Explanation:
a) since
v = vo + a*t ,
where v= velocity at time t , vo= velocity at time t=0 and a= acceleration
,then
a= (v-vo)/t
replacing values
a= (v-vo)/t = (73.14 m/s - 0 m/s)/( 30* 10⁻³ s) = 2438 m/s²
b) the distance travelled d is
v² = vo² + 2*a*d
then
d = (v² - vo²) /(2*a) = (73.14 m/s)² - 0²)/(2*2438 m/s²)= 1.0971 m
a) the acceleration is a= 2438 m/s²
b) the distance travelled during serve is d = 1.0971 m
What is acceleration?Acceleration represents the rate at which velocity should be changed with time, with respect to both speed and direction. Since acceleration contains both a magnitude and a direction, it is a vector quantity.
Calculation of acceleration & distance:a) since
[tex]v = vo + a\times t[/tex]
Here
v= velocity at time t ,
vo= velocity at time t=0
and a= acceleration
Now
[tex]a= (v-vo)\div t\\\\ =(73.14 m/s - 0 m/s)/( 30\times 10^{-3} s)[/tex]
= 2438 m/s²
b) Now the distance traveled d is
[tex]v^2 = vo^2 + 2\times a\times d \\\\d = (v^2 - vo^2) \div (2\timesa) \\\\=(73.14 m/s)^2 - 0^2)\div (2\times 2438 m/s^2)[/tex]
= 1.0971 m
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You are driving west in your car on a clear, warm day. Up ahead, you see a dark band of ominous clouds. Soon, you drive beneath the towering bank of clouds, and things get very dark. For twenty minutes, you are pelted with rain and even some hail. You keep driving west, even though it seems a little sketchy at times. On the radio, you hear that the next county to the north is under a tornado warning. Then the sky lightens, and there is only a little rain. You notice that the temperature outside has dropped by 15°F (8°C). What have you just driven across?
Answer:
You have just driven across the Cold front.
Explanation:
Weather Front:
In Meteorology, weather fronts are simply boundaries between two air masses of different densities. There are four weather fronts and one of them is cold front and other fronts are warm front, stationary front and occluded front.
Cold Front:
In Meteorology, cold front is a boundary of cold air mass that is advancing under the warm air mass.
In our scenario, as the driver was moving in the west and he listened on radio, the next country on north is under tornado warning. So, when he passed the cold front, he experienced the conditions described.
Two long straight wires enter a room through a window. One carries a current of 3.0 ???? into the room while the other carries a current of 5.0 ???? out. The magnitude in Tm of the path integral ∮ ????⃗ ∙ ???????? around the window frame is:
Answer:
[tex]\begin{equation}\\\oint_LB.dl\\\end{equation}[/tex] = -8πx[tex]10^{-7}[/tex]
Explanation:
If you need calculate
[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex]
You can use the Ampere's Law
[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex] = [tex]I_{in}[/tex]μ
Where [tex]I_{in}[/tex]: Current passing through the window
μ : Free space’s magnetic permeability
μ = 4πx[tex]10^{-7} T.m.A^{-1}[/tex]
Then
[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex] = (3-5)4πx[tex]10^{-7}[/tex]
[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex] = -8πx[tex]10^{-7}[/tex]
The magnitude of in the path integral is 2.5*10^-6 T.M
Data;
I1 = 5A
I2 = 3A
μo = 4π * 10^-7 T.m.A^-1
Ampere LawAmpere law states that the sum of the length of elements multiplied by the magnetic field in the path of the length elements is equal to the permeability multiplied by the electric current enclosed in the loop.
Mathematically;
∮B.dl = I*μo
I = current passing through the windowμo = free space magnetic permeability[tex]\int B.\delta s = \mu I\\\int B.\delta s = \mu (5A - 3A)\\\int B.\delta s = 4\pi * 10^-^7 * 2A\\\int B.\delta s = 2.5*10^-^6 T.M[/tex]
The magnitude of in the path integral is 2.5*10^-6 T.M
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A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the magnitude of the tension in the string is F. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the sting?
(A) The magnitude of the tension increases to four times its original value, 4F.
(B) The magnitude of the tension reduces to half of its original value, F/2.
(C) The magnitude of the tension is unchanged.
(D) The magnitude of the tension reduces to one-fourth of its original value, F/4.
(E) The magnitude of the tension increases to twice its original value, 2F.
When a stone is whirled at double the speed, the tension in the string increases to four times its original value, assuming the radius of the whirl remains the same.
Explanation:The tension in a string whirling a stone in a circle at a constant speed is directly proportional to the square of the speed. If the boy doubles the speed in the scenario you gave, keeping the radius of the circle unchanged, the tension in the string would increase as the square of that factor. So, between the options given, if the boy increases the speed of the stone so that it makes two complete revolutions every second instead of one, the magnitude of the tension in the string increases to four times its original value. Thus, the correct answer is (A) the magnitude of the tension increases to four times its original value, 4F.
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The tension in the string of a whirling stone increases by a factor of four when the speed of rotation doubles and the radius remains the same. It is because the tension is directly proportional to the square of the speed of the stone.
Explanation:The tension in the string of a whirling stone is related to the centripetal force, which is directly proportional to the square of the speed of rotation and the mass of the stone, and inversely proportional to the radius of the circle. If the speed of rotation doubles (from one revolution per second to two revolutions per second) and the radius of the circle remains the same, the resulting tension in the string (centripetal force) increases by a factor of four.
Hence, the answer is (A) The magnitude of the tension increases to four times its original value, 4F.
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An 80.0-g piece of copper, initially at 295°C, is dropped into 250 g of water contained in a 300-g aluminum calorimeter; the water and calorimeter are initially at 10.0°C.
What is the final temperature of the system? (Specific heats of copper and aluminum are 0.092 0 and 0.215 cal/g⋅°C, respectively. cw = 1.00 cal/g°C)
a. 12.8°C
b. 16.5°C
c. 28.4°C
d. 32.1°C
Answer:
b. 16.5°C
Explanation:
[tex]m_{c}[/tex] = mass of piece of copper = 80 g
[tex]c_{c}[/tex] = specific heat of piece of copper = 0.0920 cal/g°C
[tex]T_{ci}[/tex] = Initial temperature of piece of copper = 295 °C
[tex]m_{w}[/tex] = mass of water = 250 g
[tex]c_{w}[/tex] = specific heat of water = 1 cal/g°C
[tex]T_{wi}[/tex] = Initial temperature of piece of copper = 10 °C
[tex]m_{al}[/tex] = mass of calorimeter = 300
[tex]c_{al}[/tex] = specific heat of calorimeter = 0.215 cal/g°C
[tex]T_{ali}[/tex] = Initial temperature of calorimeter = 10 °C
[tex]T[/tex] = Final equilibrium temperature
Using conservation of heat
Heat lost by piece of copper = heat gained by water + heat gained by calorimeter
[tex]m_{c} c_{c} (T_{ci} - T) = m_{w} c_{w} (T - T_{wi})+ m_{al} c_{al} (T - T_{ali})\\(80) (0.092) (295 - T) = (250) (1) (T - 10) + (300) (0.215) (T - 10)\\T = 16.5 C[/tex]
The final temperature of the system is 16.4 ⁰C.
Conservation of energyThe final temperature of the system is determined by applying the principle of conservation of energy as shown below;
Heat lost by piece of copper = heat gained by water + heat gained by calorimeter
mCc(Tc - T) = mCw(T - Tw) + mCl(T - Tl)
80 x 0.09 x (295 - T) = 250 x 1 x (T - 10) + 300 x 0.215 x (T - 10)
2124 - 7.2T = 250T - 2500 + 64.5T - 645
5269 = 321.7T
T = 5269/321.7
T = 16.4 ⁰C
Thus, the final temperature of the system is 16.4 ⁰C.
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Spaceships A and B are traveling directly toward each other at a speed 0.5c relative to the Earth, and each has a headlight aimed toward the other ship.
What value do technicians on ship B get by measuring the speed of the light emitted by ship A's headlight?
Answer choices
1).75c
2)1.0c
3)1.5c
4) .5c
When opening a hinged door, the handle is usually on the side of the door farthest from the hinges. Rank the difficulty of opening the door if you were to push or pull on the following locations on the door:
Hardest To Open
Easiest Open
1. As close to the hinges as possible
2. the center of the door
3. The edge farthest from the hinges
The order of difficulty from the hardest to the easiest would be subject to the concept we have of Torque.
The Torque principle defines us that
[tex]\tau = Fd[/tex]
Where,
F = Force
d = Distance
As the distance increases, the force applied must be less to make the movement of the object so we have to
[tex]F \propto \frac{1}{d}[/tex]
Hence we have to be the distance inversely proportional to the force to turn the door the order would be:
As close to the hinges as possible (Hardest)> The center of the door> the edge farthest from the hinges (Easies)
1>2>3
The correct ranking of the difficulty of opening the door from hardest to easiest is as follows: 1. As close to the as possible, 2. The center of the door, 3. The edge farthest.
To understand the ranking, consider the physics of opening a door. The difficulty of opening a door is related to the torque required to rotate it around. Torque [tex](\(\tau\))[/tex] is calculated by the equation [tex]\(\tau = r \times F\)[/tex], where [tex]\(r\)[/tex] is the distance from the axis of rotation to the point where the force [tex](\(F\))[/tex] is applied.
1. As close possible: When you apply force close, the value of [tex]\(r\)[/tex] is smallest. This results in the smallest torque, meaning you have to exert more force to achieve the necessary torque to open the door, making it the hardest location to open the door.
2. The center of the door: Applying force at the center of the door increases the distance [tex]\(r\)[/tex]. This results in a larger torque for the same amount of force compared, but it is still not the most efficient point to open the door.
3. The edge farthest: The farthest edge provides the longest lever arm, meaning [tex]\(r\)[/tex] is at its maximum. This allows for the greatest torque for a given force, making it the easiest location to open the door. This is also why door handles are typically placed farthest.
A circuit consists of a coil that has a self-inductance equal to 4.3 mH and an internal resistance equal to 16 Ω, an ideal 9 V battery, and an open switch--all connected in series. At t = 0 the switch is closed. Find the time when the rate at which energy is dissipated in the coil equals the rate at which magnetic energy is stored in the coil.
Answer:
t = 186.2 μs
Explanation:
Current in LR series circuit
[tex]I(t) = I_{s}( 1 - e^{-Rt/L)}[/tex]----(1)
steady current = I_{s} = V/R
time constant = τ =[tex]L/R =4.3 * 10^{-3} / 16\\[/tex]
= 0.268 ms
magnetic energy stored in coil = [tex]U_{L} = \frac{1}{2}LI^{2}[/tex]
rate at which magnetic energy stored in coil= [tex]\frac{d}{dt}U_{L} =\frac{d}{dt} \frac{1}{2}LI^{2} \\ = LI\frac{dI}{dt}\\[/tex]----(2)
rate at which power is dissipated in R:
[tex]P = I^{2}R[/tex]---(3)
To find the time when the rate at which energy is dissipated in the coil equals the rate at which magnetic energy is stored in the coil equate (2) and (3)
[tex]I^{2}R=LI \frac{dI}{dt}[/tex]
[/tex]I=\frac{L}{R}\frac{dI}{dt}[/tex]----(4)
differentiating (1) w.r.to t
[tex]I(t)=I_{f} (1-e^{\frac{Rt}{L} })[/tex]
[tex]\frac{dI}{dt} = I_{f}\frac{d}{dt}(1-e^{\frac{-Rt}{L} } )[/tex]
[tex]\frac{dI}{dt}= I_{f}(-\frac{R}{L} e^{\frac{-Rt}{L} } )\\[/tex]---(5)
substituting (5) in (4)
[tex]I=I_{f}e^{-\frac{Rt}{L} }[/tex]----(6)
equating (1) and (6)
[tex]I_{f}( 1- e^{-\frac{Rt}{L} } ) = I_{f}e^{-\frac{Rt}{L} }[/tex]
[tex]1 - e^{-\frac{Rt}{L} } = e^{-\frac{Rt}{L} }[/tex]
[tex]\frac{1}{2}= e^{-\frac{Rt}{L} }[/tex]
[tex]t= -\frac{L}{R}ln\frac{1}{2}[/tex]
L= 4.3 mH
R= 16 Ω
t = 186.2 μs
Two moles of helium are initially at a temperature of 21.0 ∘Cand occupy a volume of 3.30×10−2 m3 . The helium first expands at constant pressure until its volume has doubled. Then it expands adiabatically until the temperature returns to its initial value. Assume that the helium can be treated as an ideal gas.B)What is the total change in internal energy of the helium?C)What is the total work done by the helium?D)What is the final volume of the helium?
Answer:
(B) The total internal energy of the helium is 4888.6 Joules
(C) The total work done by the helium is 2959.25 Joules
(D) The final volume of the helium is 0.066 cubic meter
Explanation:
(B) ∆U = P(V2 - V1)
From ideal gas equation, PV = nRT
T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3
P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal
∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules
(C) P2 = P1(V1÷V2)^1.4 =148140.4(0.033÷0.066)^1.4= 148140.4×0.379=56134.7 Pascal
Assuming a closed system
(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules
(C) Final volume = 2×initial volume = 2×0.033= 0.066 cubic meter
When he prepares food, Edgar wants to use ingredients that are not high in trans fat. Based on this information, which of the following fats should he include in his recipes?a. Stick margarine made with olive oilb. Partially-hydrogenated peanut oilc. Corn oild. Shortening
Answer:
Corn oil
Explanation:
Researchers say that corn oil is safer option than olive oil when it comes to reducing sugar levels of the blood and cholesterol levels and that it is often successful in reducing blood pressure. Corn oil is a polyunsaturated fat with some monounsaturated properties. Also, Corn oil is not high in trans fat responsible for various diseases.
If a 6-m cylindrical bar has a constant density of p= 5 g/cm for its left half and a constant density p = 6 g/cm for its right half, what is its mass?
To find the mass of the cylindrical bar, calculate the volumes of the two halves using the formula V = πr²h, where r is the radius and h is the height, and multiply them by their respective densities. Then divide the total mass by the total length of the bar.
Explanation:To find the mass of the cylindrical bar, we need to consider the densities of its left and right halves. The left half has a density of 5 g/cm, while the right half has a density of 6 g/cm. The total length of the bar is 6 m, so we can calculate the volumes of the two halves using the formula V = πr²h, where r is the radius and h is the height.
Let's assume the left half has a radius of r1 and a height of 6 m, and the right half has a radius of r2 and a height of 6 m. The total mass can then be calculated by multiplying the volume of each half by its respective density and summing the results. Finally, we divide the mass by the total length of the bar to get the mass per meter.
Select the correct statement to describe when a sample of liquid water vaporizes into water vapor
Answer:
This procces is called evaporation.
Explanation:
When you have liquid water that is transformed into steam, a phase change is called evaporation. The temperature for the evaporation of water depends on the pressure, for example for water at atmospheric pressure the temperature of evaporation is equal to 100°C. as the pressure increases are achieved evaporation temperatures higher. When that happens, the phase change temperature of the water is not increasing, as the process that takes place is the transfer of latent heat and applies only to changes of phase, that is to say at atmospheric pressure when it has 100% of the steam this will be at 101°C.
2 eagles hit each other at a 90 degree angle they grab each other, what is their velocity and direction?
Answer:
This is an example of completely inelastic collision, since they grab each other and move as a single object after the collision.
The conservation of momentum requires
[tex]\vec{P_1} + \vec{P_2} = \vec{P}_{final}[/tex]
where
[tex]\vec{P_1} = m_1v_1 \^x\\\vec{P_2} = m_2v_2 \^y[/tex]
Their final momentum is
[tex]\vec{P}_{final} = m_1v_1 \^x + m_2 v_2 \^y[/tex]
Their final velocity and direction is
[tex]\vec{v}_{final} = \frac{m_1v_1}{m_1 + m_2}\^x + \frac{m_2v_2}{m_1 + m_2}\^y[/tex]
Although not states in the question, let's consider the case that the masses and speeds of the eagles are the same:
[tex]\vec{v}_{final} = \frac{v}{2}\^x + \frac{v}{2}\^y[/tex]
A solar heating system has a 25.0% conversion efficiency; the solar radiation incident on the panels is 1 000 W/m2.
What is the increase in temperature of 30.0 kg of water in a 1.00-h period by a 4.00-m2 -area collector? (cw = 4 186 J/kg⋅°C)
a. 14.3°Cb. 22.4°Cc. 28.7°Cd. 44.3°C
Answer:
c. 28.7C
Explanation:
Since the solar radiation incident on the panel is 1000W/m2 and the collector has an area of 4m2. We can conclude that the solar power generated is
1000 * 4 = 4000 W or J/s
Within 1 hour, or 3600 seconds this solar power generator should generate (with 25 % efficiency):
E = 0.25 * 4000 * 3600 = 3600000 J or 3.6 MJ
This energy will be converted to heat to heat up water. Using water heat specific of 4186 J/kg C we can find out how much temperature has raised:
[tex]\Delta T = \frac{E}{c*m} = \frac{3600000}{30 * 4186} = 28.7^oC[/tex]
So C. is the correct answer
Students connect a spring scale to a block on a rough horizontal surface. The students use the spring scale to measure the magnitude of the horizontal force needed to pull the block at a constant speed. Which of the following statements explains why two forces exerted between objects ar equal in magnitude?
a. The gravitational and normal forces exerted on the block, because they are a Newton's third-law pair.
b. The frictional force and force exerted by the spring scale on the block, because they are a Newton's third-law pair.
c. The normal force and the frictional force exerted on the block, because objects always exert forces of equal magnitude on each other.
d. The frictional forces that the block and the surface exert on each other, because objects always exert forces of equal magnitude on each other.
For there to be a reaction there must also be action. In the horizontal movement there is a balance in which the magnitude of the Forces in opposite directions must be in total 0. The only horizontal forces are friction and the force exerted by the spring scale. For this reason the correct answer is B:
The frictional force and force exerted by the spring scale on the block, because they are a Newton's third-law pair.
A container made of steel, which has a coefficient of linear expansion 11 ✕ 10−6 (°C)−1, has a volume of 55.0 gallons. The container is filled to the top with turpentine, which has a coefficient of volume expansion of 9.0 ✕ 10−4 (°C)−1, when the temperature is 10.0°C. If the temperature rises to 25.5°C, how much turpentine (in gal) will spill over the edge of the container?
The amount of spilled turpentine due to temperature rise can be calculated by comparing the changes in volume of the steel container and the contained turpentine, using the principles and mathematical formulas of thermal expansion.
Explanation:The question asks us to determine how much turpentine, which exhibits a greater rate of thermal expansion compared to the steel container, will spill over when the temperature rises. We need to apply the principles of thermal expansion to calculate this. Both the steel container and the turpentine expand with the increase in temperature, but since turpentine has a higher coefficient of volume expansion than steel, more turpentine will expand than the container can accommodate, resulting in some turpentine spilling over.
To calculate the amount of spilled turpentine, we need to find the change in volume for both the container and turpentine, and subtract the former from the latter. The change in volume due to thermal expansion can be calculated by using the equation ΔV = βV0*(T2 - T1), where β is the coefficient of volume expansion, V0 is the initial volume, and T2 and T1 are the final and initial temperatures respectively.
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To find how much turpentine will overflow, calculate the change in volume due to thermal expansion for both the steel container and the turpentine, and subtract the change in volume of the steel from that of the turpentine.
Explanation:To solve this problem, we need to consider the thermal expansion of both the steel container and the turpentine. Thermal expansion is the increase in size of a body due to a change in temperature. This is described mathematically by the formula ΔV = βV₀ΔT, where ΔV is the change in volume, β is the coefficient of volume expansion, V₀ is the original volume, and ΔT is the change in temperature.
First, calculate the change in volume for the steel container and the turpentine separately. For the steel, β is 11 x 10^-6 (°C)^-1 and for turpentine, β is 9.0 x 10^-4 (°C)^-1. The original volume, V₀, is 55 gallons for both, and the change in temperature, ΔT, is 25.5°C - 10.0°C = 15.5°C.
Performing these calculations will give you the change in volume for the steel and the turpentine. The difference in these two volumes will tell you how much turpentine will overflow as the temperature increases.
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Air enters a turbine operating at steady state at 8 bar, 1400 K and expands to 0.8 bar. The turbine is well insulated, and kinetic and potential energy effects can be neglected. Assuming ideal gas behavior for the air, what is the maximum theoretical work that could be developed by the turbine in kJ per kg of air flow?
To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables
Mathematically this can be determined as
[tex]\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}[/tex]
Where
[tex]T_1 =[/tex]Temperature at inlet of turbine
[tex]T_2 =[/tex] Temperature at exit of turbine
[tex]P_1 =[/tex] Pressure at exit of turbine
[tex]P_2 =[/tex]Pressure at exit of turbine
The steady flow Energy equation for an open system is given as follows:
[tex]m_i = m_0 = m[/tex]
[tex]m(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W[/tex]
Where,
m = mass
[tex]m_i[/tex] = mass at inlet
[tex]m_0[/tex]= Mass at outlet
[tex]h_i[/tex] = Enthalpy at inlet
[tex]h_0[/tex] = Enthalpy at outlet
W = Work done
Q = Heat transferred
[tex]V_i[/tex] = Velocity at inlet
[tex]V_0[/tex]= Velocity at outlet
[tex]Z_i[/tex]= Height at inlet
[tex]Z_0[/tex]= Height at outlet
For the insulated system with neglecting kinetic and potential energy effects
[tex]h_i = h_0 + W[/tex]
[tex]W = h_i -h_0[/tex]
Using the relation T-P we can find the final temperature:
[tex]\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}[/tex]
[tex]\frac{T_2}{1400K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}[/tex]
[tex]T_2 = 725.126K[/tex]
From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK
So:
[tex]W = h_i -h_0[/tex]
[tex]W = C_p (T_1-T_2)[/tex]
[tex]W = 1.005(1400-725.126)[/tex]
[tex]W = 678.248kJ/Kg[/tex]
Therefore the maximum theoretical work that could be developed by the turbine is 678.248kJ/kg
Final answer:
The maximum theoretical work that could be developed by the turbine, assuming ideal gas behavior, is approximately 0.3195 kJ per kg of air flow.
Explanation:
The maximum theoretical work that could be developed by the turbine can be calculated using the ideal gas law equation:
W = Cv(Tin - Tout)
where W is the work done per unit mass, Cv is the specific heat capacity at constant volume, Tin is the inlet temperature, and Tout is the outlet temperature.
First, we need to find the value of Cv for air. For ideal gases, the specific heat capacity at constant volume can be calculated using the equation:
Cv = (R/M)
where R is the gas constant and M is the molar mass of the gas.
In this case, we will use the molar mass of air, which is approximately 28.97 g/mol. The gas constant R is 8.314 J/(mol·K).
Substituting the values into the equation, we get:
Cv = (8.314 J/(mol·K)) / (28.97 g/mol)
Cv = 0.287 J/(g·K)
Now we can calculate the work:
W = (0.287 J/(g·K))(1400 K - 288 K)
W = 0.287 J/(g·K) x 1112 K
W = 319.544 J/g
Finally, we convert the work from Joules to kilojoules:
W = 319.544 J/g * (1 kJ/1000 J)
W ≈ 0.3195 kJ/g
The circuit below shows some of the circuity in a small toy robot. When the circuit is on, the robot moves its arms (the motor) and blinks a light on its head (the light bulb).
A. How much current is in the circuit when these things are happening? Hint: To find the total resistance of the circuit, add the two resistances together.
Answer:
dear can you provide the circuit of the robot
Choose True or False for all statements
1. When the Celsius temperature doubles, the Fahrenheit temperature doubles.
2. When heat is added to a system, the temperature must rise.
3. In a bimetallic strip of aluminum and brass which curls when heated, the aluminum is on the inside of the curve.
4. An iron plate has a hole cut in its center. When the plate is heated, the hole gets smaller.
5.When the pendulum of a grandfather clock is heated, the clock runs more slowly.
6.Cold objects do not radiate heat energy.
Answer:
Explanation:
1 . When the Celsius temperature doubles, the Fahrenheit temperature doubles - FALSE
A change of temperature from 300 F to 600 F changes celsius degree from 148 to 315 °C
2. When heat is added to a system, the temperature must rise.
- FALSE
When we add heat to a system of mixture of ice and water , temperature does not rise until whole of ice melts.
3. In a bimetallic strip of aluminum and brass which curls when heated, the aluminum is on the inside of the curve
- FALSE
Coefficient of linear thermal expansion of aluminium is more so it will be on the outer side.
4. An iron plate has a hole cut in its center. When the plate is heated, the hole gets smaller.
- FALSE
hole will be bigger. There will be photogenic expansion .
5.When the pendulum of a grandfather clock is heated, the clock runs more slowly.
TRUE
Time period is directly proportional to square root of length of pendulum.
when time period increases , clock becomes slow.
6.Cold objects do not radiate heat energy
- FALSE
Object radiates energy at all temperature .
6 .
The statements are (1)False, (2)False, (3)True, (4)False, (5)True, (6)False
Here are the answers to your statements:
False. When the Celsius temperature doubles, the Fahrenheit temperature does not double. The two scales are related by the formula [tex]F = \left(\frac{9}{5} \cdot C\right) + 32[/tex]False. When heat is added to a system, its temperature does not always necessarily rise; it might cause a phase change instead.True. In a bimetallic strip of aluminum and brass, the aluminum, having a higher coefficient of thermal expansion, will indeed be on the inside of the curve when heated.False. When an iron plate with a hole is heated, thermal expansion causes both the plate and the hole to enlarge.True. When the pendulum of a grandfather clock is heated, it expands, causing the clock to run more slowly due to the increase in the pendulum's length.False. Cold objects do still radiate heat energy, although they radiate less heat energy compared to hotter objects.Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. A person can clearly see objects only if they are located between 34 cm and 180 cm away from his eyes. Bifocal lenses are used to correct his vision. What power lens (in diopters) should be used in the top half of the lens to allow him to clearly see distant objects
Answer:
The power of top half of the lens is 0.55 Diopters.
Explanation:
Since, the person can see an object at a distance between 34 cm and 180 cm away from his eyes. Therefore, 180 cm must be the focal length of the upper part of lens, as the top half of the lens is used to see the distant objects.
The general formula for power of a lens is:
Power = 1/Focal Length in meters
Therefore, for the top half of the lens:
Power = 1/1.8 m
Power = 0.55 Diopters
A 100 kg marble slab falls off a skyscraper and falls 200 m to the ground without hitting anyone. Its fall stops within milliseconds, so that there is no loss of thermal energy to its surroundings if its temperature is measured immediately after it stops. By how much has its temperature changed as a result of the fall, if we ignore energy gained or lost as a result of its interaction with the atmosphere? Cmarble = 860 J/(kg oC) 0.57 °C 1.14 °C 2.28 °C 4.56 °C
Answer:
Δ T = 2.28°C
Explanation:
given,
mass of marble = 100 Kg
height of fall = 200 m
acceleration due to gravity = 9.8 m/s²
C_marble = 860 J/(kg °C)
using conservation of energy
Potential energy = heat energy
[tex]m g h = m C_{marble}\Delta T[/tex]
[tex]g h =C_{marble}\Delta T[/tex]
[tex]\Delta T= \dfrac{g h}{C_{marble}}[/tex]
[tex]\Delta T= \dfrac{9.8 \times 200}{860}[/tex]
Δ T = 2.28°C
The temperature change which occurs as a result of the fall, if we ignore energy gained or lost is 2.28°C
What is Temperature?
This is defined as the degree of hotness or coldness of a substance.
Parameters
mass of marble = 100 Kg
height of fall = 200 m
acceleration due to gravity = 9.8 m/s²
C of marble = 860 J/(kg °C)
Using conservation of energy
mgh = mcΔT
ΔT = gh/c
= 9.8 × 200 / 860
= 2.28°C
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Penguin #1 travels at speed v=0.9 c relative to penguin #2. At the instant that both are at the origin, they synchronize their clocks to both read 1:00 AM. Penguin #1 reports that her meter stick has length one meter. If #2 were to measure the length of the meter stick of #1, what would he observe its length to be?
Answer:
L = 0.44 [m]
Explanation:
Here we can use the Lorentz transformation related to length to solve it:
[tex]L=L_{0}\sqrt{1-\beta^{2}} [/tex]
Where:
L₀ is the length of the moving reference frame (penguin #1)
L is the length of the fixed reference frame (penguin #2)
β is the ratio between v and c
We know that v = 0.9c so we can find β.
[tex]\beta = \frac{0.9c}{c}=0.9[/tex]
[tex]L=1 [m]\sqrt {1-0.9^{2}} = 0.44 [m] [/tex]
Therefore, the length of the meter stick of #1 observed by #2 is 0.44 m.
I hope it helps you!