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3 years ago

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Block A is accelerating with Block B at a rate of 0.800 m/s2 along a frictionless surface. It suddenly encounters a surface that

supplies 25.0 N a friction. What is the new Acceleration?
A. 0.4 m/s2

B. 1.0 m/s2

C. 0.8 m/s2

D. 0.6 m/s2

1 answer:

Natalija [7]3 years ago

7 0

Answer:

$a=0.6\ m/s^2$

Explanation:

The attached figure shows the whole description. Considering the applied force is 100 N.

The acceleration of both blocks A and B, $a=0.8\ m/s^2$

Firstly calculating the mass m using the second law of motion as :

F = ma

m is the mass

$m=\dfrac{F}{a}$

$m=\dfrac{100\ N}{0.8\ m/s^2}$

m = 125 kg

It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N

$(F-F')=ma$

$(100-25)=125\times a$

$a=\dfrac{75}{125}=0.6\ m/s^2$

So, the new acceleration of the block is $0.6\ m/s^2$ . Hence, this is the required solution.

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Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

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Answer:

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Explanation:

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$u_2$ = Velocity of the person before collision

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<em />

<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>

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