Answer:
8.33*10^-16 Watt
Explanation:
Given that
Length of the rod, l = 2 m,
Area of the rod, A = 2 x 2 mm² = 4*10^-6 m²
resistivity of the rod, p = 6*10^-8 ohm metre,
Potential difference of the rod, V = 0.5 V
Let R be the resistance of the rod, then
R = p * l / A
R = (6*10^-8 * 2) / (4*10^-6)
R = 3*10^14 ohm
Heat generated per second = V² / R Heat = (0.5)² / (3*10^14)
Heat = 0.25 / 3*10^14
Heat = 8.33*10^-16 Watt
Therefore, the rate at which heat is generated is 8.33*10^-16 Watt
While visiting the planet Mars, Moe leaps straight up off the surface of the Martian ground with an initial velocity of 105 m/s to catch a sleeping juicy Martian cricket. It took Moe 56 seconds to reach the bug. The gravitational acceleration on Mars is 3.75 2 m / s . Find how far the cricket was off the ground when it became Moe’s lunch. Must use integration to derive your results.
Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly shrink fit together and have to be separated. What axial force, Pa, is needed for this if the diametral interference is 0.005 mm and the coefficient of friction is 0.2? E (steel) = 207 x 103 MPa (N/mm2)
Answer:
The axial force is [tex]P = 15.93 k N[/tex]
Explanation:
From the question we are told that
The diameter of the shaft steel is [tex]d = 50mm[/tex]
The length of the cylindrical bushing [tex]L =100mm[/tex]
The outer diameter of the cylindrical bushing is [tex]D = 70 \ mm[/tex]
The diametral interference is [tex]\delta _d = 0.005 mm[/tex]
The coefficient of friction is [tex]\mu = 0.2[/tex]
The Young modulus of steel is [tex]207 *10^{3} MPa (N/mm^2)[/tex]
The diametral interference is mathematically represented as
[tex]\delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}[/tex]
Where [tex]P_B[/tex] is the pressure (stress) on the two object held together
So making [tex]P_B[/tex] the subject
[tex]P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}[/tex]
Substituting values
[tex]P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }[/tex]
[tex]P_B = 5.069 MPa[/tex]
Now he axial force required is
[tex]P = \mu * P_B * A[/tex]
Where A is the area which is mathematically evaluated as
[tex]\pi d l[/tex]
So [tex]P = \mu P_B \pi d l[/tex]
Substituting values
[tex]P = 0.2 * 5.069 * 3.142 * 50 * 100[/tex]
[tex]P = 15.93 k N[/tex]
A certain computer chip that has dimensions of 3.67 cm and 2.93 cm contains 3.5 million transistors. If the transistors are square, what must be their maximum dimensions (in terms of microns or µm)? (Note: Devices other than transistors are also on the chip, and there must be room for the interconnections among the circuit elements. Transistors smaller than 0.7 μm are now commonly and inexpensively fabricated.)
Answer:
1.56 × 10^-3 cm.
Explanation:
So, we are given the following parameters from the question above;
Length = 3.67 cm, breadth = 2.93 cm, and the number of embedded transistors = 3.5 million.
Step one: find the area of the computer chip.
Therefore, Area = Length × breadth.
Area = 3.67 cm × 2.93 cm.
Area of the computer chip = 10.7531 cm^2. = 10.75 cm^2.
Step two: find the area of one transistor
The area of one transistor is; (area of the computer chip) ÷ (number of embedded transistors).
Hence;
The area of one transistor= 10.7531/4.4 × 10^6.
The area of one transistor= 2.44 × 10^-6 cm^2.
=> Note that We have our transistors as square, therefore;
The maximum dimension = √ (2.44 × 10^-6) cm^2.
The maximum dimension= 1.56 × 10^-3 cm.
In an electric circuit, resistance and current are ____
A. directly proportional
B. inversely proportional
C. have no effect on each other
In an electric circuit, resistance and current are ____
A. directly proportional
B. inversely proportional
C. have no effect on each other
Explanation:
A
Problem 16.71 A bowler projects an 8-in.-diameter ball weighing 12 lb along an alley with a forward velocity v0 of 15 ft/s and a backspin 0 of 9 rad/s. Knowing that the coefficient of kinetic friction between the ball and the alley is 0.10, determine (a) the time t1 at which the ball will start rolling without sliding, (b) the speed of the ball at time t1, (c) the distance the ball will have traveled at time t1.
Answer:
Find the attachments for complete solution
An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10.0°C. Two metallic blocks are placed into the water. One is a 75.0-g piece of copper at 60.0°C. The other has a mass of 70.0 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20.0°C.
Determine the specific heat of the unknown sample.
Answer:
The specific heat of the unknown sample is 1822.14 J/kg.k
Explanation:
Given;
mass of aluminum calorimeter, [tex]M_c[/tex] = 100 g
mass of water, [tex]M_w[/tex] = 250 g
stabilizing temperature of water-calorimeter, ΔT = 10.0°C
mass of copper, [tex]M_c_u[/tex] = 75 g
initial temperature of copper, [tex]T_{cu}[/tex] = 60.0°C
mass of unknown sample, [tex]M_u[/tex] = 70.0 g
initial temperature of unknown sample, [tex]T_u[/tex] = 100°C.
The final temperature of the entire system, t = 20.0°C
Apply the principle of conservation of energy;
energy used to heat water and calorimeter is equal to energy released by copper and unknown sample.
[tex]Q = M_{cu}C_{cu} \delta T_{cu} + M__{u}C{_u} \delta T_u[/tex]
where;
Q is energy used to heat water and calorimeter
[tex]C_c_u[/tex] is the specific heat capacity of copper
[tex]C_u[/tex] is the specific heat capacity of unknown sample
Make [tex]C_u[/tex] subject of the formula;
[tex]C_u = \frac{Q-M_c_u C_c_u \delta T_c_u}{M_u \delta T_u} \\\\C_u = \frac{(C_wM_w +C_cM_c)\delta T-M_c_u C_c_u \delta T_c_u}{M_u \delta T _u} \\\\C_u = \frac{(4186*0.25 +900*0.1)10-0.075* 387 *40}{0.07* 80} \\\\C_u = \frac{11365 -1161}{5.6} \\\\C_u = 1822.14 \ J/kg.k[/tex]
Therefore, the specific heat of the unknown sample is 1822.14 J/kg.k
The specific heat of the unknown sample is;
c_u = 1823 J/Kg.k
We are given;
Mass of Aluminum calorimeter; m_c = 100 g = 0.1 kg
Mass of water; m_w = 250 g = 0.25 kg
Initial temperature of Calorimeter and water; T_c = T_w = 10°C = 283 K
Mass of Copper; m_cu = 75 g = 0.075 kg
Initial temperature of Copper; T_cu = 60°C = 333 K
Mass of unknown sample; m_u = 70 g = 0.07 kg
Initial temperature of unknown substance = 100°C = 373 K
Final temperature of system; T_f = 20°C = 293 K
Formula for quantity of heat is;
Q = mcΔt
where;
m is mass
c is specific heat capacity
Δt is change in temperature;
For the calorimeter and water , we have;
Q_cw = (m_w*c_w + m_c*c_c)Δt
Specific heat capacity of water is; c_w = 4186 J/Kg.K
Specific heat capacity of aluminium is 900 J/Kg.K
Thus;
Q_cw = ((0.25 * 4186) + (0.1 * 900))(293 - 283)
Q_cw = 11365 J
For the unknown sample and the piece of copper;
Q_cu,u = (m_cu*c_cu*Δt) + (m_u*c_u*Δt)
specific heat capacity of copper; c_cu = 385 J/Kg.K
Thus;
Q_cu,u = (0.075*385*(333 - 293)) + (0.07*c_u*(373 - 293))
Q_cu,u = 1155 + 5.6c_u
From conservation of energy principle;
Q_cw = Q_cu,u
Thus;
11365 = 1155 + 4.2c_u
c_u = (11365 - 1155)/5.6
c_u = 1823 J/Kg.k
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If the temperature of this balloon were to decrease suddenly, how would the balloon change?
A. Its mass would increase.
B. Its mass would decrease.
C. Its volume would increase.
D. Its volume would decrease.
Final answer:
The volume of a balloon decreases when its temperature is suddenly lowered, in accordance with Charles's Law, because the gas particles inside the balloon move less vigorously due to a decrease in their average kinetic energy.
Explanation:
When the temperature of a balloon decreases suddenly, according to Charles's Law, the volume of the gas inside the balloon also decreases, assuming that all other conditions, such as pressure and the amount of gas, remain constant. This occurs because a decrease in temperature leads to a decrease in the average kinetic energy of the gas particles, meaning they move less vigorously and occupy less space. Therefore, the correct answer to the question would be D. Its volume would decrease.
In deserts the temperatures can reach extreme values in deep of summer and winter. A steel bridge is being constructed and it is supposed to be 1234.567 m long at 233.15 K in the coldest possible winter. What will be the possible length of this bridge when the temperature rises to a possible extreme of +140.0°F in hottest part of summer? The average coefficient of linear expansion of the steel is 11.0123 × 10-6 K-1.
Answer:
The length of the bridge during the hottest part of summer is [tex]L_s = 1235.925 m[/tex]
Explanation:
From the question we are told that
The length of the steel bridge is [tex]L = 1234.567m[/tex]
The temperature for this length is [tex]T_1 = 233.15K[/tex]
The temperature at summer [tex]T_2 = + 140.0F = \frac{140 - 32}{180} *100 + 273= 333 K[/tex]
The coefficient of linear expansion is [tex]\alpha = 11.0123*10^{-6} K^{-1}[/tex]
Generally the change in length of the steel bridge is mathematically represented as
[tex]\Delta L = \alpha L \Delta T[/tex]
Substituting value
[tex]\Delta L = 11.0123*10^{-6} * 1234.567 (333-233.15)[/tex]
[tex]\Delta L = 1.3575 \ m[/tex]
The length of the bridge in summer is mathematically evaluated as
[tex]L_s = L + \Delta L[/tex]
Substituting values
[tex]L_s = 1234.567 + 1.3575[/tex]
[tex]L_s = 1235.925 m[/tex]
What is the wavelength of a sound wave traveling at 360 m/s and a
frequency of 150 Hz? *
1.0 m
2.4 m
оооо
Е Е ЕЕ
0.4 m
0.8 m
Answer:2.4m
Explanation:
Velocity=360m/s
Frequency=150Hz
Wavelength=velocity ➗ frequency
Wavelength=360 ➗ 150
Wavelength=2.4m
Final answer:
The wavelength of a sound wave traveling at 360 m/s and a frequency of 150 Hz is 2.4 m.
Explanation:
The wavelength of a sound wave can be calculated using the equation:
λ = v / ƒ
where λ is the wavelength, v is the speed of sound, and ƒ is the frequency of the sound wave. Given that the speed of sound is 360 m/s and the frequency is 150 Hz, we can substitute these values into the equation:
λ = 360 m/s / 150 Hz = 2.4 m
Therefore, the wavelength of the sound wave is 2.4 m.
mymathlab A diving pool that is 8 m deep and full of water has a viewing window on one of its vertical walls. Find the force on a window that is a%E2%80%8B circle, with a radius of 2m, tangent to the bottom of the pool.
Answer:
492.6 kN
Explanation:
Since the window as a radius of 2 m, its diameter is thus 4 m. Since the diving pool is 8 m deep, the height of water from the top of the window to the bottom of the pool is 8 - 4 = 4 m. The actual pressure acting on the window is thus
P = ρgh were ρ = density of water = 1000 kg/m³ g = 9.8 m/s² and h = 4 m
P = 1000 kg/m³ × 9.8 m/s² × 4 m = 39200 N/m².
Since P = F/A were F = force and A = area,
F = PA were A = area of window = πr² = π2² = 4π m² = 12.57 m²
F = 39200 N/m² × 12.57 m² = 492601.73 N = 492.6 kN
The force on the window of a diving pool due to the water pressure can be calculated by multiplying the pressure at the depth of the pool by the window's surface area. The pressure is found using density, gravity and depth, and the area is calculated from the radius of the circular window. The force comes out to be about 987 KN.
Explanation:This problem deals with hydrostatic pressure in a diving pool. The force on the window can be calculated by multiplying the pressure at the depth of the window by the surface area of the window.
First, we calculate the pressure at the depth of the pool using the formula P = ρgh, where P is the pressure, ρ (rho) is the density of the water (typically 1000 kg/m³ for fresh water), g is acceleration due to gravity (9.81 m/s²), and h is the depth (8m). This yields P = 1000 kg/m³ * 9.81 m/s² * 8m = 78480 Pa (Pascal).
Second, we calculate the area of the circular window. The area A of a circle is given by πr², where r is the radius. In this case, the radius r is 2m, so A = π * (2m)² = 12.57 m².
Finally, we calculate the force F by multiplying the pressure P by the area A. This gives F = P * A = 78480 Pa * 12.57 m² = 986857.6 N or approximately 987 KN.
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A current of 4.0 A is maintained in a single circular loop having a circumference of 80 cm. An external magnetic field of 2.0 T is directed so that the angle between the field and the plane of the loop is 20°. Determine the magnitude of the torque exerted on the loop by the magnetic forces acting upon it. Group of answer choices 0.27 N ⋅ m 0.41 N ⋅ m 0.38 N ⋅ m 0.77 N ⋅ m 0.14 N ⋅ m
The torque on a single circular loop in an external magnetic field can be calculated using the formula T = NIAB sin 0. We need to find the area of the loop using the given circumference and then substitute the values of the current, magnetic field strength, area, and angle into the formula to find the torque.
Explanation:To calculate the magnitude of the torque exerted on the circular loop by the magnetic forces, we need to use the formula for torque on a current-carrying loop in a uniform magnetic field, which is T = NIAB sin 0, where N represents the number of turns in the loop, I represents the current, A represents the area of the loop, B represents the magnetic field strength, and 0 represents the angle between the field and the plane of the loop.
In this case, since there is only a single loop, N is equal to 1. The current I is given as 4.0 A. The magnetic field strength B is given as 2.0 T. The angle 0 is 20°. The area of the loop A can be calculated from the circumference given as 80 cm or 0.8 m. Recall that the circumference of a circle is given by the formula 2πr. If the circumference C is given by 80 cm or 0.8 m, the radius r can be found by dividing the circumference by 2π. Once you've found the radius r, the area of the circle is πr².
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The correct answer is "0.14 N.m". The final torque value is 0.14 N·m on the loop.
To determine the torque exerted on the loop, we use the formula:
Torque (τ) = nIBA sin(θ)
First, find the area (A) of the loop. The circumference (C) of the loop is given by 80 cm, which is 0.80 m. From the circumference, we can find the radius (r):
C = 2πr -> r = C / 2π = 0.80 m / 2π ≈ 0.127 m
Now, calculate the area (A) of the loop:
A = πr² = π(0.127 m)² ≈ 0.0507 m²
Next, calculate the torque:
τ = nIBA sin(θ) = 1 × 4.0 A × 2.0 T × 0.0507 m² × sin(20°)
Using the value of sin(20°) ≈ 0.342:
τ ≈ 1 × 4.0 A × 2.0 T × 0.0507 m² × 0.342 ≈ 0.14 N · m
Therefore, the magnitude of the torque exerted on the loop is 0.14 N·m.
g A simple pendulum (consisting of a point mass suspended by a massless string) on the surface of the earth has a period of 1.00 s. On a distant planet, a pendulum of identical length but twice the mass is found to have the same period. What is true about the acceleration due to gravity on the distant planet? [Assume both pendulums operate free of any friction or air resistance force]
Answer:
Explanation:
The formula for time period of a pendulum is given as follows :
T = 2π[tex]\sqrt{\frac{l}{g} }[/tex]
l is length of pendulum and g is acceleration due to gravity .
So time period of pendulum is not dependent on the mass of the pendulum . If time period is same and length is also the same then acceleration due to gravity will also be the same . Hence the acceleration due to gravity at distant planet will be same as that on the earth.
A 1.20 kg copper rod rests on two horizontal rails 0.78 m apart and carries a current of 45 A from one rail to the other. The coefficient of static friction between rod and rails is 0.61. What is the magnitude of the smallest magnetic field that puts the rod on the verge of sliding
Answer:
B = 0.204T
Explanation:
To find the value of the magnetic force you use the following formula:
[tex]F_B=ILBsin\theta[/tex]
I: current of the copper rod = 45A
B: magnitude of the magnetic field
L: 0.78m
you assume that magnetic field B and current I are perpendicular between them.
The magnetic force must be, at least, equal to the friction force, that is:
[tex]F_{f}=F_{B}\\\\\mu N=\mu Mg=ILB\\\\B=\frac{\mu Mg}{IL}[/tex]
M: mass of the rod = 1.20kg
μ: coefficient of static friction = 0.61
g: gravitational acceleration constant = 9.8m/s^2
By replacing the values of the parameters you obtain:
[tex]B=\frac{(0.61)(1.20kg)(9.8m/s^2)}{(45A)(0.78m)}=0.204T[/tex]
Answer:
The magnitude of the smallest magnetic field is [tex]B = 0.1744 \ T[/tex]
Explanation:
From the question we are told that
The mass of the copper is [tex]m = 1.20 kg[/tex]
The distance of separation for the rails is [tex]d = 0.78 \ m[/tex]
The current is [tex]I = 45 A[/tex]
The coefficient of static friction is [tex]\mu = 0.61[/tex]
The force acting along the vertical axis is mathematically represented as
[tex]F = mg - F_y[/tex]
Where [tex]F_y[/tex] is the force acting on copper rod due to the magnetic field generated this is mathematically represented as
[tex]F_y = I * d * B_1[/tex]
The magnetic field here is acting towards the west because according to right hand rule magnetic field acting toward the west generate a force acting in the vertical axis
So the equation becomes
[tex]F = mg - I * d * B_1[/tex]
Here [tex]B_1 = B sin \theta[/tex]
[tex]F = mg - I * d * Bsin(\theta )[/tex]
The in the horizontal axis is mathematically represented as
[tex]F_H = ma + F_x[/tex]
Since the rod is about to move it acceleration is zero
Now [tex]F_x[/tex] is the force acting in the horizontal direction due to the magnetic field acting downward this is because a according to right hand rule magnetic field acting downward generate a force acting in the horizontal positive horizontal direction. this mathematically represented as
[tex]F_H = 0 + I * d * B_2[/tex]
So the equation becomes
[tex]F_H = I * d * B_2[/tex]
Here [tex]B_2 = B cos \theta[/tex]
[tex]F_H = I * d * Bcos (\theta)[/tex]
Now the frictional force acting on this rod is mathematically represented as
[tex]F_F = \mu * F[/tex]
[tex]F_F = \mu * (mg -( I * d * Bsin(\theta )))[/tex]
Now when the rod is at the verge of movement
[tex]F_H = F_F[/tex]
So [tex]I * d * Bcos (\theta) = \mu * (mg -( I * d * Bsin(\theta )))[/tex]
=> [tex]B = \frac{\mu mg }{I * d (cos \theta + \mu (sin \theta ))}[/tex]
Now [tex]\theta[/tex] is the is the angle of the magnetic field makes with the vertical and the horizontal and this can be mathematically evaluated as
[tex]\theta = tan^{-1} (\mu )[/tex]
Substituting value
[tex]\theta = tan^{-1} ( 0.61 )[/tex]
[tex]\theta = 31.38^o[/tex]
Substituting values into the equation for B
[tex]B = \frac{0.61 (1.20) (9.8)}{(45) (0.78) (cos (31.38) + 0.61 (sin (31.38)) )}[/tex]
[tex]B = 0.1744 \ T[/tex]
An oven mitt is made of a material that is a poor thermal conductor.
Which is the most likely reason that the material makes a good oven mitt?
A: it is made mostly of water
B: it lacks electrons
C: it is made of tightly packed particles
D: its electrons do not move freely
Answer:
D
Explanation:
The most likely reason that a material that is a poor thermal conductor makes a good oven mitt is because its electrons do not move freely.
Explanation:The most likely reason that a material that is a poor thermal conductor makes a good oven mitt is because its electrons do not move freely.
In a good thermal conductor, such as metal, the electrons are able to move freely and transfer heat quickly. However, in a poor thermal conductor, such as the material of an oven mitt, the electrons are not able to move freely, which means they cannot transfer heat easily and thus provide insulation.
Therefore, option D: its electrons do not move freely is the most likely reason that a material makes a good oven mitt.
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A horizontal rope is tied to a 57.0 kg kg box on frictionless ice. What is the tension in the rope if: You may want to review (Pages 135 - 137) . Part A The box is at rest? Express your answer as an integer and include the appropriate units. T T = nothing nothing SubmitRequest Answer Part B The box moves at a steady v x vxv_x = 4.20 m/s m/s ? Express your answer as an integer and include the appropriate units. T T = nothing nothing SubmitRequest Answer Part C The box v x vxv_x = 4.20 m/s m/s and a x axa_x = 5.80 m/ s 2 m/s2 ?
Answer:
Explanation:
A ) The surface is frictionless . If the box is at rest , there will be no tension in the rope.
B ) In this case , the box is moving with steady rate of 4.20 m /s . In this case also no force is acting on the box so tension in the rope will be nil.
C ) In this case the box is moving with acceleration of 5.8 m /s² so force on the box = mass x acceleration
= 57 x 5.8
= 330.6 N .
So tension in the rope will be equal to force acting on the rope . Hence tension = 330.6 N .
Final answer:
The tension in the rope tied to a 57.0 kg box on frictionless ice is 0 N when the box is at rest, 0 N when moving at a steady velocity of 4.20 m/s, and 331 N when the box is accelerating at 5.80 m/s².
Explanation:
When considering the tension in the rope connected to the 57.0 kg box on frictionless ice, different scenarios will result in different tensions:
Part A: Box at Rest
In the first scenario where the box is at rest, the tension in the rope will be zero Newtons (0 N). This is because, on frictionless ice, there is no other force opposing the box's motion that the rope would need to counteract.
Part B: Box Moving at Steady Velocity
For the second part, where the box is moving at a constant velocity (4.20 m/s), the tension is still zero Newtons. Even if the box is in motion, the lack of frictional forces on the ice means no net force is required to maintain the box's constant velocity.
Part C: Box Accelerating
In the third case, where the box is moving at 4.20 m/s and also accelerating at 5.80 m/s², we need to apply Newton's second law, F = ma. The tension (T) in the rope is calculated as the product of the mass (m) of the box and its acceleration (a). Thus:
T = m × a = 57.0 kg × 5.80 m/s² = 330.6 N
Therefore, the tension in the rope would be 331 N (rounded to the nearest integer).
n open rectangular tank is filled to a depth of 2 m with water (density 1000 kg/m3). On top of the water there is a 1 m deep layer of gasoline (density 700 kg/m3). The width of the tank is 1 m (the direction perpendicular to the paper). The tank is surrounded by air at atmospheric pressure. Calculate the total force on the right wall of the tank, and specify its direction. The acceleration of gravity g = 9.81 m/s2.
Answer:
gasoline zone P_net = 6860 Pa
Water zone P_net = 26460 Pa
Force direction is out of tank
Explanation:
The pressure is defined
P = F / A
F = P A
let's write Newton's equation of force
F_net = F_int - F_ext
P_net A = (P_int - P_ext) A
The P_ext is the atmospheric pressure
P_ext = P₀
the pressure inside is
gasoline zone
P_int = P₀ + ρ' g h'
water zone
P_int = P₀ + ρ' g h' + ρ_water h_water
we substitute
Zone with gasoline
P_net = ρ' g h'
P_net = 700 9.8 1
P_net = 6860 Pa
Water zone
P_net = rho’ g h’ + rho_water g h_water
P_net = 6860 + 1000 9.8 2
P_net = 26460 Pa
To find the explicit value of the force, divide by a specific area.
Force direction is out of tank
Weather balloons are high altitude balloons that are capable of reaching an altitude of 40 km (or 25 miles). They are usually filled with hydrogen or helium gas. One such balloon is filled with gas that is initially at a temperature of 300 K and at atmospheric pressure. This particular balloon can expand to a maximum diameter of 32 m when it reaches an altitude where the pressure and temperature are 0.028 atm and 190 K respectively. Determine the diameter of the balloon at lift off
Answer:
[tex]D_{1}=11.32 m[/tex]
Explanation:
We will need to use the ideal gas equation. The equation is given by:
[tex]PV=nRT[/tex]
P is the pressureV is the volumen is the amount of moleculesR is the ideal gas constantAs we have the same amount of molecules in the initial and final steps, therefore we can do this:
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex] (1)
- P(1) is the atmospheric pressure (P(1) = 1 atm) and P(2) is 0.028 atm
- T(1) is 300 K and T(2) is 190 K
- V(1) is the volume of the balloon in the first step, we can consider a spherical geometry so:
[tex]V_{1}=\frac{4}{3}\pi (\frac{D_{1}}{2})^{3}[/tex] (2)
[tex]V_{2}=\frac{4}{3}\pi (\frac{D_{2}}{2})^{3}[/tex] (3)
- D(2) = 32 m
So [tex]V_{2}=17157.3 m^{3}[/tex]
Let's solve the equation (1) for V(1)
[tex]V_{1}=\frac{T_{1}P_{2}V_{2}}{P_{1}T_{2}}[/tex]
[tex]V_{1}=\frac{300*0.028*17157.3}{1*190}[/tex]
[tex]V_{1}=758.53 m^{3}[/tex]
And using the equation (2) we can find D.
[tex]D_{1}=2(\frac{3}{4}V_{1})^{1/3}[/tex]
[tex]D_{1}=2(\frac{3}{4\pi}*758.53)^{1/3}[/tex]
[tex]D_{1}=11.32 m[/tex]
I hope it helps you!
What is the structure's center of gravity? (b) Calculate the structure's moment of inertia. (c) What is the angular acceleration experienced by the masses? (d) Determine the structures angular velocity after 10 s. (e) After 30 s, the external force is removed. What is the total energy of the system?om/homework-help/questions-and-answers/far-space-5-kg-ball-10-kg-ball-connected-10-m-long-rigid-massless-rod-steady-force-10-n-ap-q50426001
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
A car has a momentum of 20,000 kg • m/s. What would the car’s momentum be if its velocity doubles? 10,000 kg • m/s 20,000 kg • m/s 40,000 kg • m/s 80,000 kg • m/s
Answer:40000kg•m/s
Explanation:
momentum=20000kg•m/s
Since they velocity is doubled
Momentum would also be doubled, therefore momentum would be 40000kg•m/s
Answer:
C: 40,000 kg · m/s
Explanation:
I got it right on edg :)
What is the unit for force?
Answer:
N / NEWTONS
Explanation:
Named after Isaac Newton, the man who discovered gravity
Answer:
newtons and the symbol is N
Explanation:
The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on, the initial current decreases until the filament reaches its steady-state temperature. The temperature coefficient of resistivity of the filament is 4 times 10-3 K-1. The final current through the filament is one- eighth the initial current. What is the change in temperature of the filament
Answer:
The change in temperature is [tex]\Delta T = 1795 K[/tex]
Explanation:
From the question we are told that
The temperature coefficient is [tex]\alpha = 4 * 10^{-3 }\ k^{-1 }[/tex]
The resistance of the filament is mathematically represented as
[tex]R = R_o [1 + \alpha \Delta T][/tex]
Where [tex]R_o[/tex] is the initial resistance
Making the change in temperature the subject of the formula
[tex]\Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ][/tex]
Now from ohm law
[tex]I = \frac{V}{R}[/tex]
This implies that current varies inversely with current so
[tex]\frac{R}{R_o} = \frac{I_o}{I}[/tex]
Substituting this we have
[tex]\Delta T = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ][/tex]
From the question we are told that
[tex]I = \frac{I_o}{8}[/tex]
Substituting this we have
[tex]\Delta T = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ][/tex]
=> [tex]\Delta T = \frac{1}{3.9 * 10^{-3}} (8 -1 )[/tex]
[tex]\Delta T = 1795 K[/tex]
The lamp filament's temperature rise can be calculated using the equations for resistance and the current. Given the temperature coefficient of resistivity and the ratio of initial and final currents, the change in temperature is calculated to be 875K.
Explanation:The problem is solved by using the formula for the change in resistance due to temperature, R = R0(1 + αΔT), where R is the final resistance, R0 is the initial resistance, α is the temperature coefficient of resistivity, and ΔT is the change in temperature. We can relate the initial and final currents (I and If) using Ohm’s law: V = I*R = If Rf, where V is the constant voltage. Given that If = I/8, and substituting from R = R0(1 + αΔT), we get that ΔT = [(8 - 1)/(α*7)] = 875K.
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A spring hangs from the ceiling with an unstretched length of x0=0.45 m . A m1=7.9 kg block is hung from the spring, causing the spring to stretch to a length x1=0.74 m.Find the length x2of the spring when a m2=3.3 kg block is hung from the spring. For both cases, all vibrations of the spring are allowed to settle down before any measurements are made.
To find the new length of the spring when a different mass is hung from it, calculate the spring constant using the initial mass, and then use Hooke's Law to solve for the extension caused by the new mass. Add this extension to the original unstretched length of the spring to find the stretched length.
Explanation:The student's question involves finding the new length of a spring x2 when a block with mass m2 of 3.3 kg is hung from it. Given an initial unstretched length (x0) and a stretched length (x1) with block m1, we can solve for the spring constant k using Hooke's Law, which states F = k * x, where F is the force, k is the spring constant, and x is the extension from the natural length of the spring.
First, we calculate the extension caused by m1: extension_m1 = x1 - x0 = 0.74 m - 0.45 m = 0.29 m. The force exerted by m1 is F1 = m1 * g, where g is the acceleration due to gravity (9.81 m/s2). Calculating the force, we get F1 = 7.9 kg * 9.81 m/s2. Then, the spring constant k can be found using k = F1 / extension_m1.
With the value of k, we can find the extension caused by m2: extension_m2 = F2 / k, where F2 = m2 * g. Finally, the length x2 of the spring with m2 hung from it is x2 = x0 + extension_m2.
What measurement stays the same at any point in a parallel circuit?
A. current
B. voltage
C. resistivity
D. power
Answer:Voltage
Explanation:
Voltage stays the same at any point in a parallel circuit
The amount of light entering a camera or your eye is regulated by
a. an eyepiece
c. an aperture.
b. the cornea.
d. a set of compound lenses.
Answer:
c. an aperture
Explanation:
Aperture: It relates to the size of the opening, like a doorway, through which light moves into the eye, camera lens or a telescope. In human eye aperture is known as pupil, the black part in the center of the eye. The size of the pupil can increase or decrease depending upon the amount of light available. The same thing happens with a camera as well. The amount of light passing through the lens can be varied by varying the size of the aperture.
What can you infer about the prefix "thermo-" from the
word "thermoregulation?"
A
It refers to the brain
reser to the brain
It refers to temperature
It refers to feedback
It refers to nerves and hormones
Answer:
it refers to temperature I did test with the same question and got it correct.
Explanation:
A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.600c. The pursuit ship is traveling at a speed of 0.800c relative to Tatooine, in the same direction as the cruiser. (a) For the pursuit ship to catch the cruiser, should the velocity of the cruiser relative to the pursuit ship be directed toward or away from the pursuit ship? (b) What is the speed of the cruiser relative to the pursuit ship?
Answer:
a) toward the pursuit
b) 0.384c
Explanation:
a) The velocity of the cruiser relative to the pursuit should be toward the pursuit.
b) To find the speed of the cruiser relative to the pursuit ship you use the following formula:
[tex]u'=\frac{u-v}{1-\frac{uv}{c^2}}[/tex]
v: velocity of the cruiser as seen by Tatooine
u: velocity of the pursuit ship as seen by Tatooine
c: speed of light
By replacing the values of the parameters you obtain:
[tex]u'=\frac{0.800c-0.600c}{1-\frac{(0.800)(0.600)c^2}{c^2}}=0.384c[/tex]
When accounting for relativistic velocity addition, the velocity of the cruiser relative to the pursuit ship should be directed towards the pursuit ship for it to be 'caught up' with. The cruiser's relative speed is about -0.429c from the perspective of the pursuit ship.
Explanation:From the perspective of the observer on Tatooine, the cruiser is moving away from the planet at a speed of 0.600c and the pursuit ship is moving in the same direction at a speed of 0.800c. To figure out these velocities in a relative sense, we must use the theory of relativity, specifically the concept of relativistic velocity addition. In classical physics, velocities simply add or subtract, but this isn't the case when dealing with speeds close to the speed of light.
(a) For the pursuit ship to catch up with the cruiser, the velocity of the cruiser relative to the pursuit ship should be directed towards the pursuit ship.
(b) When solving for the relative speed of the cruiser from the perspective of the pursuit ship, we use the formula for relativistic velocity addition: V'=(v-u)/(1-(uv/c^2)). In this case, v is the speed of the cruiser (0.600c), u is the speed of the pursuit spacecraft (0.800c), and c is the speed of light. Plugging in the values, you will find that the speed of the cruiser relative to the pursuit ship is approximately -0.429c (where the minus sign indicates the cruiser is moving towards the pursuit ship relative to the pursuit ship's frame of reference).
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7. Which one of the following statements best describes the concept of the electric field? a) The electric field is a vector quantity that is the reaction force of electrons. b) The electric field at each point is the potential energy of a test charge divided by the amount of the test charge. c) The electric field is a distribution of vectors at points due to the presence of one or more charged objects. d) The electric field is a scalar quantity related to the total amount of charge on one or more charged objects. e) The electric field is a scalar field, which has a magnitude at each given point, similar to the temperature or pressure field.
Answer:(c)
Explanation:
The electric field is a distribution is a distribution of vectors at point due to the presence of one or more charged objects.
If two or more charged particles are present in a system then the net electric field at a point is the vector addition of all the charges.
For example if a negative and a positive charge is present then electric field of negative charge is towards the negative charge while it is away for positive charge.
option d and e are wrong as Electric field is a vector quantity
An eighteen gauge copper wire has a nominal diameter of 1.02mm. This wire carries a constant current of 1.67A to a 200w lamp. The density of free electrons is 8.5 x 1028 electrons per cubic metre. Find the magnitude of:
i. The current density ii. The drift velocity
The current density of the wire is approximately[tex]2.04 x 10^6 A/m²[/tex]elocity of the electrons in the wire is approximately [tex]1.51 x 10^-3 m/s.[/tex]
Explanation:To find the magnitude of the current density, we need to calculate the cross-sectional area of the wire and divide the current by that area. The current density (J) is given by J = I/A, where I is the current and A is the cross-sectional area. Since the wire is cylindrical, we can use the formula for the area of a circle, A = πr², where r is the radius of the wire.
Given that the diameter of the wire is [tex]1.02mm[/tex]radius can be calculated as [tex]r = (1.02mm)/2 = 0.51mm = 0.00051m[/tex]
Using this value, the cross-sectional area is [tex]A = π(0.00051m)² = 8.18 x 10^-7 m².[/tex]
Dividing the current of[tex]1.67A[/tex]e cross-sectional area, we get the current density [tex]J = (1.67A)/(8.18 x 10^-7 m²) ≈ 2.04 x 10^6 A/m².[/tex]
To find the drift velocity, we can use the formula v_d = J/(nq), where n is the density of free electrons and q is the charge of an electron. Given that the density of free electrons is [tex]8.5 x 10^28 electrons/m³[/tex]
the charge of an electron is [tex]q = 1.60 x 10^-19 C[/tex]
we can substitute these values into the formula: [tex]v_d = (2.04 x 10^6 A/m²) / (8.5 x 10^28 electrons/m³ * 1.60 x 10^-19 C/electron) ≈ 1.51 x 10^-3 m/s.[/tex]
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The current density in the wire is approximately 2.04 × 10^6 A/m^2, and the drift velocity of the electrons is approximately 1.51 × 10^-4 m/s.
Sure, let's break down the calculation step by step:
Given data:
Diameter of the wire, d = 1.02 mm = 1.02 × 10^-3 m
Current, I = 1.67 A
Power of the lamp, P = 200 W
Density of free electrons, n = 8.5 × 10^28 electrons/m^3
Calculate the cross-sectional area of the wire:
Radius, r = d/2 = 1.02 × 10^-3 m / 2 = 0.51 × 10^-3 m
Cross-sectional area, A = π * r^2
A = π * (0.51 × 10^-3 m)^2 ≈ 8.19 × 10^-7 m^2
Calculate the current density (J):
Current density is the current per unit area.
J = I / A
J = 1.67 A / 8.19 × 10^-7 m^2 ≈ 2.04 × 10^6 A/m^2
Calculate the charge carrier density (n):
Given density of free electrons, n = 8.5 × 10^28 electrons/m^3
Calculate the drift velocity (v_d):
Drift velocity is given by the formula: J = n * e * v_d
Where e is the charge of an electron (approximately 1.6 × 10^-19 C)
Rearranging for drift velocity: v_d = J / (n * e)
Substituting values: v_d = (2.04 × 10^6 A/m^2) / (8.5 × 10^28 electrons/m^3 * 1.6 × 10^-19 C)
v_d ≈ 1.51 × 10^-4 m/s
Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected by a tube that contains an ideal turbine (a mechanical device) and a valve. Initially, the valve is closed. Container A (volume VA = 2 m3) is filled with n = 1.2 moles of an ideal monatomic gas, and container B (volume VB = 3.5 m3) is empty (vacuum). When the valve is opened, gas flows between A and B, and the turbine is used to generate electricity. What is the maximum amount of work than can be done on the turbine as the gas reaches equilibrium? Remember: U = 1.5 nRT + const S = nR ln(V) + f(U,n)
Answer:
The maximum amount of work is [tex]W = 1563.289 \ J[/tex]
Explanation:
From the question we are told that
The temperature of the environment is [tex]T = 280\ K[/tex]
The volume of container A is [tex]V_A = 2 m^3[/tex]
Initially the number of moles is [tex]n = 1.2 \ moles[/tex]
The volume of container B is [tex]V_B = 3.5 \ m^3[/tex]
At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as
[tex]W = P_A V_A ln[ \frac{V_B}{V_A} ][/tex]
Now from the Ideal gas law
[tex]P_A V_A = nRT[/tex]
So substituting for [tex]P_A V_A[/tex] in the equation above
[tex]W = nRT ln [\frac{V_B}{V_A} ][/tex]
Where R is the gas constant with a values of [tex]R = 8.314 \ J/mol[/tex]
Substituting values we have that
[tex]W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ][/tex]
[tex]W = 1563.289 \ J[/tex]
The maximum amount of work done on the turbine can be found by calculating the change in internal energy of the gas as it reaches equilibrium. The work done on the turbine is given by the difference in pressure between the initial and final states of the gas, multiplied by the change in volume. In this case, the maximum work done on the turbine is (nRT/VA)(VB - VA).
Explanation:When the valve is opened and the gas flows from container A to container B, the gas will expand and reach equilibrium. Since the process is isothermal, the temperature remains constant at T = 280K. To find the maximum work done on the turbine, we need to calculate the change in internal energy of the gas using the equation ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the gas, and W is the work done on the gas. For an ideal monatomic gas, the internal energy is given by U=1.5nRT, where n is the number of moles, R is the gas constant, and T is the temperature. In this case, we have n = 1.2 moles, V = 2m3 for container A, and V = 3.5m3 for container B. Therefore, the total volume is V = VA + VB = 5.5m3. Using the ideal gas law PV = nRT, we can find the initial pressure of the gas in container A, which is P = nRT/VA. Since the gas expands to fill both containers at equilibrium, the final pressure is P = nRT/V. The work done on the turbine is given by W = P(VB - VA). Plugging in the values, we find that the maximum work done on the turbine is W = (nRT/VA)(VB - VA).
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In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 2.2 rev/s in 8.8 s. You lift the lid of the washer and notice that the tub accelerates and comes to a stop in 20.0 s. Assuming that the tub rotates with constant angular acceleration while it is starting and stopping, determine the total number of revolutions undergone by the tub during this entire time interval.
Answer:
[tex]n_{T} = 31.68\,rev[/tex]
Explanation:
The angular acceleration is:
[tex]\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}[/tex]
[tex]\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}[/tex]
And the angular deceleration is:
[tex]\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}[/tex]
[tex]\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}[/tex]
The total number of revolutions is:
[tex]n_{T} = n_{1} + n_{2}[/tex]
[tex]n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}[/tex]
[tex]n_{T} = 31.68\,rev[/tex]