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2 years ago

Answer the attached question.No Spam!⚔️⚔️⚔️

Physics

1 answer:

RoseWind [281]2 years ago

6 0

<h3>Question:5</h3>

The given data can be written in following way in coordinates axes.

(0,0), (2,2), (4,4), (6,4), (8,4), (10,6),(12,4), (14, 2).

a) Average velocity for first 4 seconds

Average velocity = Total Displacement/ Time taken

= (4-0)/(4 - 0)

=4/4

= 1/1

b) Average velocity for 4 to 8 seconds

Average velocity = Total Displacement/ Time taken

= (4 - 4)/(8-4)

= 0/4

c) Average velocity for last 6 seconds

Last 6 seconds = from 8 to 14 seconds

Average velocity = Total Displacement/Time taken

= (2 - 4)/(14 - 8)

= -2/6

<h3>Question 6:</h3>

The given data can be written in following way in coordinates axes.

(0,0), (10,20), (20,20), (30,20), (40,0)

a) State the kind of motion from Os to 10s and from 30s to 40s

It is obvious from the graph that the velocity between Os to 10s has been increased from 0 m/s to 20m/s. Hence there is a uniform Acceleration in the body.

b) What is the velocity of the body after 10s and 40s ?

It is clear from the graph and table as well that,

<h3>Velocity after 10s is 20m/s</h3>

and

<h3>Velocity after 40s is 0 m/s</h3>

c) Calculate the distance covered by the body between 10s to 30s.

Distance covered by the body between 10s to 30s will be given by area of rectangle ABCD

Area of rectangle ABCD= (30 - 10) × (20 - O)

=20×20

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Answer:25.61 m/s

Explanation:

Given

truck is moving eastbound with a velocity of 16 m/s

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Magnitude of relative velocity is

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A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. Wha

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A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. The change in the electric potential energy of the proton-field system when the proton travels to x = 2.50m is -3.40 × 10⁻¹⁶ J (Option B)

<h3 /><h3>How is the change in electric potential energy of the proton-field system calculated?</h3>

Work done on the proton =Negative of the change in the electric potential energy of the proton field
In the given case, W = -qΔV
-W = qΔV
= qEcosθ
Therefore, work done on the proton = -e(8.50× $10^2$ N/C)(2.5m)(1)
= -3.40× $10^-^1^6$ J
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Therefore the positive sign shows that the potential energy increases when the proton does the work.
The negative sign shows that the potential energy decreases when the proton does the work.