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n200080 [17]
3 years ago
6

A 10,000kg space ship is orbiting the moon with a radius of 25km from the ship to the center of the moon. It's tangential speed

is 125m/s. What is the ships centripetal acceleration?
Is this the correct set-up for this problem.

ac=10,000kg(125m/s)2/25,000m

True or False?
Physics
1 answer:
galina1969 [7]3 years ago
3 0

Answer:

False

Explanation:

ac = v^2/r

acceleration is not dependent on the mass of the orbiting object.

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How is the Sun related to nuclear electromagnetic and heat energy
Snezhnost [94]

-- The source of most of the energy that radiates from the sun is nuclear energy.

-- Most of the energy that radiates from the sun is electromagnetic energy.

-- Heat energy is part of the electromagnetic energy that radiates from the sun.
Other parts include radio, microwave, visible light, ultraviolet, and X-ray energy.

5 0
3 years ago
A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H
Aleks04 [339]

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .

Explanation:

Dont report my answer please

3 0
3 years ago
What type of energy transfer occurs through light or electromagnetic waves?
Free_Kalibri [48]
Radiation is the transfer of heat energy through space by electromagnetic radiation. Most of the electromagnetic radiation that comes to the earth from the sun is in the form of visible light. Light is made of waves of different frequencies.
4 0
3 years ago
Read 2 more answers
Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
sergeinik [125]

v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)

\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0
3 years ago
A rock is thrown down from the top of a cliff with a velocity of 3.61 m/s (down). The cliff is 28.4 m above the ground. Determin
Katarina [22]
I’m not sure look it up in the internet!
4 0
2 years ago
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