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tester [92]
3 years ago
9

A 48.9 kg meteor is moving in outer space. If a 8.6 N force is applied opposite the direction of motion, what is the deceleratio

n (in outer space we assume no friction)
Physics
1 answer:
nalin [4]3 years ago
3 0

Answer:

The deceleration is 0.18 m/s²

Explanation:

Hi there!

Using Newton´s second law, we can calculate the deceleration:

∑F = m · a

Where:

∑F = the sum of all forces in a given direction.

m = mass of the object.

a = acceleration.

Solving for a:

∑F/m = a

The only force acting on the meteor is the applied force of 8.6 N. So, the acceleration will be:

8.6 N / 48.9 kg = a

a = 0.18 m/s²

The deceleration is 0.18 m/s² or, in other words, the acceleration is -0.18 m/s²

Have a nice day!

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A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar
kakasveta [241]

Answer:

W=4037.36\ J

Explanation:

Given:

mass of ice melted, m=3.3\times10^{-2}\ kg

time taken by the ice to melt, t=5\ min=300\ s

latent heat of the ice, L=3.34\times 10^5\ J

Now the heat rejected by the Carnot engine:

Q_R=m.L

Q_R=0.033\times 3.34\times 10^5

Q_R=11022\ J

Since we have boiling water as hot reservoir so:

T_H=373\ K

The cold reservoir is ice, so:

T_L=273\ K

Now the efficiency:

\eta=1-\frac{T_L}{T_H}

\eta=1-\frac{273}{373}

\eta=26.81\%

Now form the law of energy conservation:

Heat supplied:

Q_S-W=Q_R

where:

Q_S= heat supplied to the engine

Q_S-\eta\times Q_S=Q_R

Q_S(1-\eta)=Q_R

Q_S=\frac{11022}{1-0.2681}

Q_S=15059.36\ J

Now the work done:

W=Q_S-Q_R

W=15059.36-11022

W=4037.36\ J

8 0
3 years ago
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