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Gnesinka [82]
3 years ago
7

A newly discovered planet has a radius twice as large as earth's and a mass five times as large. What is the free-fall accelerat

ion on its surface?
Physics
1 answer:
fomenos3 years ago
8 0

Answer:

free-fall acceleration = 12.25 m/s²

Explanation:

Formula for free fall acceleration is given by the gravity equation;

g = Gm/r²

Where;

M is mass of the Earth

r is radius of the Earth

G is the gravitational constant

Now, that equation applies to the earth.

For this new planet, we are told that the mass = 5 × mass of earth

And radius = 2 × radius of earth.

Thus, free fall acceleration for this planet is;

g_p = G(5m)/(2r)²

g_p = (5/4)(Gm/r²)

Gravitational constant has a value of 6.674 × 10^(−11) N.m²/kg²

Mass of the earth = 5.972 × 10^(24) kg

Radius of the earth = 6378 km = 6378000

Thus;

g_p = (5/4)(6.674 × 10^(−11) × 5.972 × 10^(24))/(6378000²) = 12.25 m/s²

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Using information about natural laws, explain why some car crashes produce minor injuries and others produce catastrophic injuri
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3 years ago
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Explain in your own words the interaction between the electric and magnetic fields that make up a light wave.
iVinArrow [24]

Answer:

They oscillates perpendicularly to one another, the oscillation of one field generates the other field.

Explanation:

In a light wave, an oscillating electric field of a light wave produces a magnetic field, and the magnetic field also oscillates to produce an electric field. The magnetic field and the electric field of a light wave both oscillates perpendicularly to one another. The resultant energy and direction of the wave generated as a result of these oscillating fields is propagated perpendicularly to both fields.

8 0
3 years ago
Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 10
NNADVOKAT [17]

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

      m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v

Putting the given values into the above formula as follows.

      m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v

     2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v

           -4.12 \times 10^{3} kg m/s = (10^{4} kg) v

                   v = \frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}

                      = -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.

8 0
3 years ago
How much heat must be added to make a 5g substance with a specific heat of 2 J/gC that has its temperature go up 10 degrees? Q =
zlopas [31]

Answer:

100 Joule

Explanation:

Amount of heat in agiven body is given by Q = m•C•ΔT

where m is the mass of the body

c is the specific heat capacity of body. It is the amount of heat stored in 1 unit weight of body which raises raises the temperature of body by 1 unit of temperature.

ΔT is the change in the temperature of body

___________________________________________

coming back to problem

m = 5g

C = 2J/gC

since, it is given that temperature of body increases by 10 degrees, thus

ΔT = 10 degrees

Using the formula for heat as given

Q = m•C•ΔT

Q = 5* 2 * 10  Joule= 100 Joule

Thus, 100 joule heat must be added to  a  5g substance with a specific heat of 2 J/gC to raise its temperature go up by 10 degrees.

8 0
3 years ago
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