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klio [65]
3 years ago
12

A particle is moving with a velocity of 60.0 m/s in the positive x direction at t= 0. Between t= 0 and t= 15.0 s the velocity de

creases uniformly to zero. What was the acceleration during this 15.0-s interval? What is the significance of the sign of your answer?
Physics
1 answer:
Juliette [100K]3 years ago
7 0

Answer:

Acceleration of the object is given as

a = - 4 m/s^2

so here negative sign shows that the direction of acceleration is opposite to the velocity of the object

Explanation:

As we know that velocity decreases uniformly from maximum to zero

so here in this uniform deceleration we know that

v_i = 60 m/s

v_f = 0

t = 15 s

now by using the equation of kinematics we have

v_f - v_i = at

0 - 60 = a(15)

a = - 4 m/s^2

so here negative sign shows that the direction of acceleration is opposite to the velocity of the object

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A capacitor has a capacitance of 0. 40 µF at a voltage of 9. 0 V. What is the charge on each plate of the capacitor? µC.
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The capacitor is a device that can store electrical energy. It is a two-conductor configuration. The charge on each plate of the capacitor will be 3.6 µC.

<h3>What is a capacitor?</h3>

A capacitor is a device that can store electrical energy. It is a two-conductor configuration separated by an insulating medium that carries charges of equal size and opposite sign.

An electric insulator or vacuum, such as glass, paper, air, or a semi-conductor termed a dielectric, can be used as the non-conductive zone.

The given data in the problem is;

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V is the  voltage = 9. 0 V

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The formula for the capacitor is given as;

\rm Q=CV \\\\ \rm Q=0. 40 \times 9. 0 \\\\ \rm Q=3.6 \ \mu C.

Hence the charge on each plate of the capacitor will be 3.6 µC.

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brainly.com/question/14048432

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Answer:

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