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Iteru [2.4K]
3 years ago
9

a proton is fired with a speed of 200,000m/s from the midpoint of the capacitor toward the positive plate. This speed is insuffi

cient to reach the positive plate. What is the proton's speed as it collides with the negative plate

Physics
2 answers:
NemiM [27]3 years ago
3 0

Answer: The proton speed = 3 × 10^5m/s

Explanation: The electric P.E change the proton if It can reach the positive plate.

The workdone

Pavlova-9 [17]3 years ago
3 0

I have attached an image of the diagram showing the nature of this motion

Answer:

Protons speed = 2.96 x 10^(5) m/s

Explanation:

A) At closest point of approach to the positive plate, the proton came to rest momentarily.

Thus;

Loss in Kinetic Energy = Gain in Electric potential energy

Hence;

(1/2)(mv^(2)) = eΔV

So, ΔV = (mv^(2))/(2e)

Mass of proton = 1.673 × 10-27 kilograms

Proton elementary charge(e) = 1.6 x 10^(-19) coulumbs

And from the question v = 200,000 m/s

So, ΔV = [1.673 × 10^(-27) x 200000^(2)] / (2 x 1.6 x 10^(-19)) = 209 V

This is less than 250V which is half of the charge at the positive plate shown in the diagram.

Therefore, the speed is insufficient to reach the positive plate from P to Q.

B) Gain in KE = qΔV

Thus; 1/2mvf^(2) - 1/2mvi^(2) = eΔV

Where, vf is final velocity and vi is initial velocity.

So simplifying, we get;

vf^(2) - vi^(2) = (2eΔV)/m

So, vf = √[(2eΔV)/m) + (vi^(2))

= √[(2 x 1.6 x 10^(-19) x 250)/(1.673 × 10^(-27)) + (200,000^(2))

= 2.96 x 10^(5) m/s

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2 years ago
Answer it pls!!!!!!!!!!!
Archy [21]

Answer:

Fractional error = 0.17

Percent error = 17%

F = 112 ± 19 N

Explanation:

Plug in the values to find the force:

F = (3.5 kg) (20 m/s)² / (12.5 m) = 112 N

Find the fractional error:

ΔF/F = Δm/m + 2Δv/v + Δr/r

ΔF/F = 0.1/3.5 + 2(1/20) + 0.5/12.5

ΔF/F = 0.17

Multiply by 100% to find the percent error:

ΔF/F × 100% = 17%

Solve for the absolute error:

ΔF = 0.17 × 112 N = 19 N

Therefore, the force is:

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8 0
3 years ago
A house is lifted from its foundations onto a truck for relocation. The house is pulled upward by a net force of 2850 N. This fo
Gwar [14]

Answer:

m = 95000 kg

Explanation:

Given that,

Net force acting on the house, F = 2850 N

Initial speed, u = 0

Final speed, v = 15 cm/s = 0.15 m/s

We need to find the mass of the house. Let the mass be m. We know that the net force is given by :

F = ma

Where

a is the acceleration of the house.

So,

F=m\dfrac{v-u}{t}\\\\m=\dfrac{Ft}{(v-u)}\\\\m=\dfrac{2850\times 5}{(0.15-0)}\\\\m=95000\ kg

So, the mass of the house is equal to 95000 kg.

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2 years ago
An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid
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Answer:

a)

- 7.04\times10^{11} J

b)

5.03\times10^{5} N

Explanation:

a)

m = mass of the asteroid = 43000 kg

v_{o} = initial speed of asteroid = 7600 m/s

v = final speed of asteroid = 5000 m/s

W = Work done by the force on asteroid

Using work-change in kinetic energy theorem

W = (0.5) m (v^{2} - v_{o}^{2})\\W = (0.5) (43000) (5000^{2} - 7600^{2})\\W = - 7.04\times10^{11} J

b)

F = magnitude of force on asteroid

d = distance traveled by asteroid while it slows down = 1.4 x 10⁶ m

Work done by the force on the asteroid to slow it down is given as

W = - F d\\- 7.04\times10^{11} = - F (1.4\times10^{6})\\F = 5.03\times10^{5} N

4 0
3 years ago
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