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ivolga24 [154]
3 years ago
10

Did changing the angle of the incline affect the following parameters?

Physics
1 answer:
MArishka [77]3 years ago
4 0
Increasing the level of an incline:
Increases final velocity
Increases the work done
Increases the initial potential energy
Increases the final kinetic energy
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An object with a mass of 2 kilograms is accelerated at 5 m/s2.
monitta

Answer:

It would be 40

Explanation:

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(HELP ASAP) A student’s cell phone battery is almost dead, and she plugs it into an electrical outlet to charge. As it charges,
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Explanation:

The total energy of the system remains conserved. It just converts from one form of energy to another.

The battery of the cell phone contains chemicals. As the electricity is passed through the cell phone, the electrical energy is converted to chemical energy. As she notices that the phone is warm and after the battery has been fully charged and the student unplugs her cell phone, the phone vibrates when a message arrives, this shows the conversion of chemical energy to mechanical energy, thermal energy etc.

Hence, we can say that the correct option is (b).

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Human activities that impact the water cycle
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d

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A number of human activities can impact on the water cycle: damming rivers for hydroelectricity, using water for farming, deforestation and the burning of fossil fuels.

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What is the order of the electromagnetic spectrum from highest to lowest energy?
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<span> Radio waves, microwaves, infra red, visible light, ultra violet, x-rays, gamma rays. </span>

<span />

4 0
3 years ago
The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5
nexus9112 [7]

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7.55 km/s

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The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

G\frac{mM}{R^2}=m\frac{v^2}{R}

where

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M=5.97\cdot 10^{24} kg is the mass of the Earth

R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s

6 0
3 years ago
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