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Aliun [14]
2 years ago
6

A. 24.89 B. 25.89 C. 17.74 D. 19.73

Physics
1 answer:
Veronika [31]2 years ago
5 0

Answer: D

Explanation:

Just did it got an 100

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 <span>Nuclei of u-238 atoms are </span><span>unstable and spontaneously emit alpha particles.</span>
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Why is it a good idea to start with room temperature water in the calorimeter?
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It is a good idea to start with room temperature water in the calorimeter because the room temperature water helps to determine the heating up/cooling down because of the environment as the experiment takes place. Because the calorimeter heat is the same as the heat of the water.

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Maggie's truck decelerates from 18.00 m/s to rest in 3.30 s. If the Maggie's mass is 100.0 kg and the truck is 1810 kg, calculat
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look at my Explanation

Explanation:

If the Maggie's mass is 100.0 kg and the truck is 1810 kg, calculate the magnitude of the net (unbalanced) force that can cause the acceleration.

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Find the magnitude of the force at point x = 15cm if k=10500 N/m when the work done by this force is WD= 1/2k x2
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3 years ago
A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou
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Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

<u>Mathematical expression for the Newton's second law of motion is given as:</u>

F=\frac{dp}{dt} ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

p=m.v

Now, equation (1) becomes:

F=\frac{d(m.v)}{dt}

<em>∵mass is constant at speeds v << c (speed of light)</em>

\therefore F=m.\frac{dv}{dt}

and, \frac{dv}{dt} =a

where: a = acceleration

\Rightarrow F=m.a

also

F\propto \frac{1}{dt}

so, more the time, lesser the force.

<em>& </em><u><em>Impulse:</em></u>

I=F.dt

I=m.a.dt

I=m.\frac{dv}{dt}.dt

I=m.dv=dp

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0
2 years ago
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