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3 years ago

If a 50 N force pulls on a 10 kg object,how much acceleration will occur?

Chemistry

2 answers:

Zinaida [17]3 years ago

6 0

Answer:

The value of acceleration occurred is $5 m/s^2$ .

Explanation:

Force is a push or pull on an object with mass which causes in change in its velocity Or it is the product of its mass and acceleration by which that object is moving.

Force = $mass\times acceleration$

F = ma

We have:

F = 50 N, a = ? , m = 10 kg

Putting in the values we get,

$50 N=10 kg\times a$

$a=\frac{50 N}{10 kg}=5 m/s^2$

The value of acceleration occurred is $5 m/s^2$ .

meriva3 years ago

4 0

Assuming that the force of friction does not equal to the force at which you are pulling at, hence Fnet is not zero, to solve for acceleration need to use the following equation:

Fnet = ma
a = Fnet/m
a = 50 N/10 kg
a = 5 N/kg or a = 5 m/s^2.

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Answer:

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3 years ago

Read 2 more answers

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Leni [432]

Answer:

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Explanation:

The entropy change can be calculated using the following equation:

$\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})$

Where:

Q: is the energy transferred = 5.0 MJ

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$T_{i}$ : is the initial temperature = 1000 K

$T_{f}$ : is the final temperature = 500 K

Hence, the entropy change is:

$\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}$

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!

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