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3 years ago

A stone is dropped from a cliff and falls 9.44 meters. What is the speed of the stone when it reaches the ground?

Physics

1 answer:

Lunna [17]3 years ago

3 0

Answer:

option A

Explanation:

given,

height of the drop of stone = 9.44 m

speed of the stone = ?

As the stone is dropped the energy of the stone will be conserved.

using conservation of energy.

Potential energy = Kinetic energy

$m g h = \dfrac{1}{2} m v^2$

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8 \times 9.44}$

$v = \sqrt{185.024}$

v = 13.60 m/s

Hence, the correct answer is option A

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A 63.3 kg wood board is resting on very smooth ice in the middle of a frozen lake. A 35.7 kg boy stands at one end of the board.

Vlada [557]

The velocity of the board relative to the ice is zero, since both are at rest.

<h3>What is relative velocity?</h3>

Relative velocity is the velocity of an object in relation to another reference object or point.

When two objects are travelling or moving with the same velocity in the same direction, the relative velocity one relative to the other is zero.

Also, when two objects are at rest, the relative velocity one relative to the other is zero.

Therefore, the velocity of the board relative to the ice is zero, since both are at rest.

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2 years ago

The atmospheric pressure at the surface of titan is.

Harrizon [31]

Answer:

about 1.5 bars

Titan's atmosphere is similar to Earth's both in the predominance of nitrogen gas and in surface pressure, which is about 1.5 bars, or 50 percent higher than sea-level pressure on Earth.

Explanation:

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2 years ago

A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the

sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

Since the frictional force $f_k$ is equal to $f_k=\mu_k N=\mu_k mg$ , then we have that the acceleration is:

$a=\frac{\mu_k mg}{m}=\mu_k g$

Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$