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katrin [286]
3 years ago
14

he "escape velocity from Earth (the speed required to escape Earth's gravity) is 2.5 x 10 miles per hour. What is this speed in

m/s? (1 mile 1609 m)
Physics
1 answer:
Natasha_Volkova [10]3 years ago
4 0

Answer: 11.17\ \text{ m/s}

Explanation:

Given : The escape velocity : v=2.5\times10\text{ miles per hour}

We know that 1 mile = 1609 meters  (approx)

and 1 hour= 3600 seconds

To convert escape velocity 2.5 x 10 miles per hour into m/s , we need to multiply it by 1609.34 and divide it by 3600.

Thus, the escape velocity in m/s is given by :-

v=2.5\times10\times\dfrac{1609}{3600}\\\\=11.1736111111\approx11.17\text{ m/s}

Hence, the speed in m/s = 11.17

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Describe two sources of earth's energy that are not produced
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What color could be created by mixing cyan and yellow pigment?
Cloud [144]

Answer:

green pigment

Explanation:

green pigment

3 0
2 years ago
Read 2 more answers
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
A cylindrical storage tank has a radius of 1.35 m. When filled to a height of 3.45 m, it holds 14,014 kg of a liquid industrial
Kruka [31]

Answer:

709.93 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³

From the question above,

D = m/v.............................. Equation 1

Where D = density, m = mass, v = volume.

Note: The volume of the liquid is equal to the volume of the height occupied by the liquid in the container

Since the tank is cylindrical,

v = πr²h........................ Equation 2

Where r = radius of the the tank, h = height of the liquid in the tank

Substitute equation 2 into equation 1

D = m/(πr²h)............... Equation 3

Given: m = 14014 kg, r = 1.35 m, h = 3.45 m, π = 3.14

Substitute into equation 3

D = 14014/(3.14×1.35²×3.45)

D = 14014/19.74

D = 709.93 kg/m³

6 0
3 years ago
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