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anastassius [24]
3 years ago
5

Part C: Quantitative Problems when vf is not 0

Physics
1 answer:
Alina [70]3 years ago
3 0

Answer:

(a)

\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s

(b)

1120 N

Explanation:

Change in velocity, \triangle v is given by subtracting the initial velocity from the final velocity and expressed as \triangle v= v_f -v_i

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

\triangle v=0-8=-8\ m/s

To find m\triangle v then we substitute 7 kg for m and -8 m/s for \triangle v therefore \triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s

(b)

The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie

a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}

Substituting t with 0.05 s then a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.05}=-160 m/s^{2}

Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N

Therefore, the impact force is equivalent to 1120 N

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A paper filled capacitor is charged to a potential difference of 2.1 V and then disconnected from the charging circuit. The diel
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Part b)

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Q = kC(2.1)

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Part b)

Now the capacitor plates are again isolated and unknown dielectric is inserted between the plates

So again charge is same so potential difference is given as

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aev [14]

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

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third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

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\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

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5 0
3 years ago
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