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DENIUS [597]
3 years ago
12

When a box of chocolate bars is in mechanical equilibrium, what can be correctly said about all the forces that act on it?

Physics
1 answer:
rjkz [21]3 years ago
3 0

Answer:

All the forces are opposite and equal and will give net force zero.

Explanation:

Mechanical equilibrium is the equilibrium in which the total force on the system is zero means the system is neither accelerated nor any kind of torque on the system.

The mechanical system can also be defined as the equal forces are applied in opposite direction in a system which cancels out all the forces, will give net force zero.

Therefore, when a box of chocolate bars is in mechanical equilibrium all the forces in this system  are equal and the opposite which balances each other will give net force zero.

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We need to verify the thrid equation of kinematics here

\\ \tt\longmapsto v^2-u^2=2as

\\ \tt\longmapsto 20^2=2(-8)s

\\ \tt\longmapsto 400=-16s

\\ \tt\longmapsto s=|400/-16|

\\ \tt\longmapsto s=25m

The squirrel has a good luck ,Car gets stopped just 1m away from the squirrel .

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A permanent magnet is pushed into a wire, left there for a while, and then pulled out. During which time does a current run thou
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C. while the magnet is moving

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Electromagnetic induction implies the production of electric current by mere movement of a magnet with respect to a coil or wire.

In the given question, current would be induced in the wire only when the magnet moves. That is either when the magnet is pushed into a wire, or when pulled out. But no current would flow through the wire when the magnet is left there for a while.

The current is induced because of the motion involved. Thus, the appropriate option is C.

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What purpose does a female pine cone serve?
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The answer is <span>It protects the seed until it matures.

Through the process of elimination:
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Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

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v_2(t)=124\dfrac{\rm m}{\rm s}-gt

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1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

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{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

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Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

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That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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