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aev [14]
2 years ago
11

A test charge gains 10 joules of potential energy as it moves through an electric field. It starts its movement at point 1 and e

nds its movement at point 2. The potential difference between the two points is 4 volts. How many coulombs of charge are carried by the test charge
Physics
1 answer:
Nataliya [291]2 years ago
8 0

2.5 C coulombs of charge are carried by the test charge.

<h3>What is test charge?</h3>

A test charge is a charge of such a modest size that it has little impact on the field surrounding the site where it is placed. To determine the strength and direction of an electric field, a positive test charge is a unit positive charge.

U = qV

where,

U = 10 j

V = 4V

then, q = U/V

q = 10/4

q = 2.5 C

to learn more about test charge go to - brainly.com/question/21830343

#SPJ4

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A proton with a velocity in the positive x-direction enters a region where there is a uniform magnetic field B in the positive y
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Negative z-direction.

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4 years ago
Compare and contrast microscopic and macroscopic energy transfer Give least three comparisons for each
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3 0
3 years ago
A truck covers 40.0 m in 7.45 s while uniformly slowing down to a final velocity of 3.50 m/s.
astra-53 [7]

Answer:

(A) Original velocity will be 7.244 m /sec

(B) Acceleration will be  -0.5026m/sec^2

Explanation:

We have given distance covers by truck s = 40 m

Time taken by truck to cover this distance t = 7.45 sec

Final velocity v = 3.50 sec

According to second equation of motion

S=ut+\frac{1}{2}at^2

40=u\times 7.45+\frac{1}{2}\times a\times 7.45^2

7.45u+27.751a=40-----eqn 1

According to first equation of motion

v = u + at

So 3.5=u+7.45a-----eqn2

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a = -0.5026m/sec^2

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6 0
4 years ago
A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a spe
Rasek [7]

Answer:

56.5\ \text{s}

21.13\ \text{m/s}

Explanation:

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

s = Displacement

Here the kinematic equations of motion are used

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}

Time the car is at constant velocity is 39 s

Time the car is decelerating is 5 s

Total time the car is in motion is 12.5+39+5=56.5\ \text{s}

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v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}

s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}

v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2

s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}

The total displacement of the car is 156.25+975+62.5=1193.75\ \text{m}

Average velocity is given by

\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}

The average velocity of the car is 21.13\ \text{m/s}.

6 0
3 years ago
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