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Anastasy [175]
3 years ago
10

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

Physics
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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Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

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Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

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Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

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So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
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8 0
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Answer:

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