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3 years ago

A 5 kg rock is dropped down a vertical mine shaft. How long does it take to reach the bottom, 79 meters below?

Physics

1 answer:

likoan [24]3 years ago

7 0

Answer:

The time for the rock to reach the bottom is 4.02 seconds.

Explanation:

Given;

mass of the rock, m = 5 kg

height of the rock fall, h = 79 meters

The time to drop to the given height is given by;

$t = \sqrt{\frac{2h}{g} }$

where;

t is the time to fall to the bottom

g is acceleration due to gravity = 9.8 m/s²

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*79}{9.8} }\\\\t = 4.02 \ s$

Therefore, the time for the rock to reach the bottom is 4.02 seconds.

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nevsk [136]

Answer:

The mass is $m = 3.75 \ kg$

Explanation:

From the question we are told that

The initial temperature is $T_1 = 15^oC$

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$H = 0.35 * E$

=> $H = 0.35 * 9000 00$

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Converting to kcal

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3 years ago

The famous black planet, haunch, has a radius of 106 m, a gravitational acceleration at the surface of 4 m/s2 , and the tangenti

levacccp [35]

Gravity on the surface = 4 m/s^2
Now, the acceleration due to centripetal motion, a = v^2/R

Where,
v= 10^3 m/s, R = 10^6 m
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3 years ago

A 2 kg object with a weight of 20 N is being pulled up by a rope with a tension of 12N what is the acceleration of the object

son4ous [18]

Answer:

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

Explanation:

Given;

mass of the object, m = 2 kg

weigh of the object, W = 20 N

tension on the rope, T = 12 N

The acceleration of the object is calculated by applying Newton's second law of motion as follows;

T = F + W

T = ma + W

ma = T - W

$a = \frac{T-W}{m} \\\\a = \frac{12 - 20}{2} \\\\a = -4 \ m/s^2$ (the negative sign indicates deceleration of the object)

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

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3 years ago