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2 years ago

an object is dropped from the top of a building which is 75 meters high. what's it's velocity before it reaches ground

Physics

1 answer:

Nataly [62]2 years ago

3 0

Hi there! :)

$\large\boxed{38.34 m/s}$

Use the following kinematic equation to solve:

$v_{f}^{2} = v_{i}^{2} + 2ad$

Where:

vf = final velocity (?)

vi = initial velocity (0 m/s, object started at rest)

a = acceleration (gravity = 9.8 m/s)

d = distance travelled (75 m)

Plug in the given values into the equation:

$v_{f}^{2} = 0^{2} + 2(9.8)(75)$

Simplify:

$v_{f}^{2} = 2(9.8)(75)\\\\v_{f}^{2} = 1470$

Take the square root of both sides:

$v_{f} = \sqrt{1470} = 38.34 m/s$

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2 years ago

What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface

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Answer:

171.5 N

Explanation:

The gravitational force on an object due to the Earth is given by

$F=mg$

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m is the mass of the object

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The acceleration due to gravity at a certain height h above the Earth is given by

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where:

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$M=5.98\cdot 10^{24} kg$ is the Earth's mass

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So the acceleration due to gravity is

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.37\cdot 10^6 + 6.38\cdot 10^6)^2}=2.45 m/s^2$

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2 years ago

26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the

denpristay [2]

Answer:

c. 1600J

Explanation:

The loss in potential energy of the boy is given by:

$U=mg \Delta h$

where

m = 40 kg is the mass of the boy

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h = 4 m + 0.02 m = 4.02 m$ is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)

Substituting, we find

$\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J$

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2 years ago

What are the assumptions on which the cosmological principle is based?.

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Hope this helps!
Please give Brainliest!

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2 years ago

(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro

Kipish [7]

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:

$v = \frac{m + M}{m} \cdot \sqrt{2gh}$

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum. </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:

$\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}$ (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively. </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}$

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}$

$\frac{v_{2}}{v_{1}} = 1.41$

Therefore, the second projectile was 1.41 times faster than the first.

I hope it helps you!

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2 years ago