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Lera25 [3.4K]
3 years ago
11

A body is dropped from the roof of a 20 m high building by how much:

Physics
1 answer:
USPshnik [31]3 years ago
7 0

Answer:

t = 2.01 s

Vf = 19.7 m/s

Explanation:

It's know through the International System that the earth's gravity is 9.8 m/s², then we have;

Data:

  • Height (h) = 20 m
  • Gravity (g) = 9.8 m/s²
  • Time (t) = ?
  • Final Velocity (Vf) = ?

==================================================================

Time

Use formula:

  • \boxed{t=\sqrt{\frac{2*h}{g}}}

Replace:

  • \boxed{t=\sqrt{\frac{2*20m}{9.8\frac{m}{s^{2}}}}}

Everything inside the root is solved first. So, we solve the multiplication of the numerator:

  • \boxed{t=\sqrt{\frac{40m}{9.8\frac{m}{s^{2}}}}}

It divides:

  • \boxed{t=\sqrt{4.08s}}

The square root is performed:

  • \boxed{t=2.01s}

==================================================================

Final Velocity

use formula:

  • Vf = g * t

Replace:

  • Vf = 9.8 m/s² * 2.01 s

Multiply:

  • Vf = 19.7 m/s

==================================================================

How long does it take to reach the ground?

Takes time to reach the ground in <u>2.01 seconds.</u>

How fast does it hit the ground?

Hits the ground with a speed of <u>19.7 meters per seconds.</u>

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Answer:

The minimum coefficient of friction required is 0.35.  

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

-F_{f} + F = 0      

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\mu mg = ma

\mu = \frac{a}{g}

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

v_{f}^{2} = v_{0}^{2} + 2ad

a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}

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v_{0}: is the initial speed of the truck = 17.9 m/s

a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}        

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.  

\mu = \frac{a}{g}  

\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}

\mu = 0.35

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!

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<u>Unlike theory</u>, which is the set of rules and principles that describe and explain a particular phenomenon and <u>is subject to changes as new evidence emerges that gives meaning to it.  </u>

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Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl
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Answer:

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