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Free_Kalibri [48]
3 years ago
12

If a 5 kg bowling ball is lifted to a height of 2 meters, how much potential

Physics
1 answer:
natka813 [3]3 years ago
8 0

Answer:

\boxed {\boxed {\sf 98.1 \ Joules}}

Explanation:

The formula for potential energy is:

E_p=mgh

where m is the mass, g is the gravitational acceleration, and h is the height.

The mass is 5 kilograms and the height is 2 meters. Assuming this ball is on Earth, then the gravitational acceleration is 9.81 meters per square second.

m= 5 \ kg \\g= 9.81 \ m/s^2 \\\h= 2 \ m

Substitute the values into the formula.

E_p= ( 5 \ kg )(9.81 \ m/s^2)( 2 \ m)

Multiply the first two numbers.

E_p=49.05 \ kg*m/s^2 (2 \ m )

Multiply again.

E_p= 98.1 \ kg*m^2/s^2

  • 1 kilogram square meter per square second is equal to 1 Joule.
  • Our answer is equal to 98.1 Joules

E_p= 98.1 \ J

The ball has 98.1 Joules of potential energy.

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Hoochie [10]
Convert the units of power,W = 7 hp = 7 * 745.69 = 5219.83 WCalculate the power input to the pump using the efficiency of the pump equationn=Wpump/Wshaft
Substitute 0.82 for n and 5219.83 for Wshaft0.82=Wpump/5219.83   Wpump=0.82*5219.83=4280.26 WCalculate the mass flow ratein=Wpump/(gz_2 )Where g is the acceleration due to gravity, and z_2 is the elevation of water. Substitute 4280.26 for Wpump, 9.81 m/s^2 for g, and 19m for z_2in = 4280.26 / 9.81 * 19 = 22.9640 m^3/sCalculate the volume flow rate of waterV=m/ρWhere ρ is the density of water. Substitute 22.9640  m^3/s for in and 1000  m^3/kg for ρ, we get V = 22.9640 / 1000 = 0.0230 kg/sTherefore, the volume flow rate of water is 0.0230 kg/s
7 0
3 years ago
How fast would you say the earth rotates near the equator?
PtichkaEL [24]
460 meters per second, or about 1,000miles per hour.
8 0
3 years ago
Which of the statements concerning light are true? The speed of light is the same no matter what material it is traveling throug
wel

Answer:

The statements that are true concerning light are the last three statements:

  • Its propagation direction is perpendicular to both the electric field and the magnetic field.
  • It moves at a constant speed through a vacuum.
  • The speed of light in matter is less than the speed of light in a vacuum.

Explanation:

<em>Light</em> is <em>electromagnetic waves.  </em>

The properties of the electromagnetic waves were established by James Clerk Maxwell.

They included that they are the result of the oscillation of a <em>magnetic field </em>in phase with an <em>electric field</em> which are always is always <em>perpendicular</em> to each other.

Also, the electromagnetic waves propagate at right-angles to the direction of both the magnetic and the electric field,  meaning that they are a type of transverse wave.

Thus, the second statement (<em>"Its propagation direction is parallel to both the electric field and the magnetic field"</em>) is false, and the fourth statement ("Its propagation direction is perpendicular to both the electric field and the magnetic field") is true.

On the other hand, it is a postulate of the special theory of relativity that the speed of light is a constant (absolute value) in vacuum: nothing can travel faster than what light travels in vacuum. Thus, the fifth statement, <em>"It moves at a constant speed through a vacuum"</em> is true.

About the speed of light in matter, it is always less than the speed of light in vacuum. Thus, the first statement, "<em>the speed of light is the same no matter what material it is traveling through</em>", and the third statement "<em>the speed of light in matter is greater than the speed of light in a vacuum"</em> are false; while the last statement, "<em>the speed of light in matter is less than the speed of light in a vacuum</em>" is true.

The explanation on why the speed of light is less in a medium than in vacuum is related with the fact that at nanoscopic level the waves suffer polarization which means deviations from the straighi path, which makes that the net straight propagation is slower.

8 0
3 years ago
A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.
kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight=\sqrt{\frac{2H}{g}}

substituting value

  • =\sqrt{\frac{2*25}{9.8}}
  • =2.26seconds

<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u*\sqrt{\frac{2H}{g}}

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Answer:

Here are 5:

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Hope that was helpful.Thank you!!!

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