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joja [24]
3 years ago
4

A stationary block has a charge of +6.0×10−8 C. A 0.80-kg cart with a charge of +4.0×10−8 C is initially at rest and separated f

rom the block by 0.4 m. The cart is released and moves along a frictionless surface to a distance of 1.0 m from the block. Consider the objects as point charges.
A) Determine the electric potential energy of the initial state.
B)Determine the electric potential energy of the final state.
C)Determine the change in electric potential energy.
D)Determine the final speed of the cart. Assume that the change in electric potential energy is fully transformed into kinetic energy of the cart.
Physics
1 answer:
Mama L [17]3 years ago
3 0

Answer:

(a) U = 5.40 × 10^-7 J

(b) U = 2.16 × 10^-7 J

(c) ∆ U = 3.24 × 10^-7 J

(d) v = 9.0 × 10^-4 m/s

Explanation:

Electric Potential Energy, U = kq1q2/d

K = 9 × 10^9 Nm²/C², q1 = 6.0 × 10^-8 C, q2 = 4.0 × 10^-8 C, d = 0.4 m

(a) U = (9 × 10^9 × 6.0 × 10^-8 × 4.0 × 10^-8) / 0.4

U = 5.40 × 10^-7 J

(b) At d = 1.0 m,

U = (9 × 10^9 × 6.0 × 10^-8 × 4.0 × 10^-8) / 1.0

U = 2.16 × 10^-7 J

(c) ∆ U = (5.40 × 10^-7) - (2.16 × 10^-7)

∆ U = 3.24 × 10^-7 J

(d) ∆ U = ½mv²; where m = 0.80 Kg, ∆ U = 3.24 × 10^-7 J

v = √(2 × ∆ U / m)

v = √(2 × 3.24 × 10^-7 / 0.8)

v = 9.0 × 10^-4 m/s

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2 years ago
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, h
Alex73 [517]

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

5 0
2 years ago
Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the
Westkost [7]

Answer:

They are able to balance torques due to gravity.

F_1 L_1 = F_2L_2

Explanation:

When two friends of different masses will balance themselves on see saw then at equilibrium position the see saw will remain horizontal

This condition will be torque equilibrium position where the see saw will not rotate

Here we can say

F_1 L_1 = F_2L_2

here we know that force is due to weight of two friends

and their positions are different with respect to the lever about which see saw is rotating

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2 years ago
A heat engine (Power Cycle) with a thermal efficiency of 35 percent efficiency produces 750 kJ of work. Heat transfer to the eng
frosja888 [35]

Answer:

a) The schematic illustrating is attached

b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ

c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ

Explanation:

b) The heat transfer to the engine and the heat transfer from the engine to the air is:

Q_{1} =\frac{W}{n}

Where

W = 750 kJ

n = 35% = 0.25

Replacing:

Q_{1} =\frac{750}{0.35} =2142.86kJ

Q_{2} =Q_{1} -W=2142.86-750=1392.86kJ

c) The efficiency of Carnot engine is:

n=1-\frac{300K}{550K} =0.455

The heat transfer to the heat engine is:

Q_{1c} =\frac{750}{0.455} =1648.35kJ

The heat transfer from the heat engine is:

Q_{2c} =1648.35-750=898.35kJ

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Differences between Pressure and upthrust​
Angelina_Jolie [31]

Answer:

Pressure is equal to the ratio of thrust to the area in contact. Upthrust is a force exerted by the fluids on an object placed in the fluid . Upthrust acts in upward direction.

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