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fiasKO [112]
2 years ago
8

We are going to focus on the gravitational force between Earth and our Moon. The Earth has a mass of about 6 x 10/24 kg and the

Moon has a
mass of about 7.3 x 10/22 kg. They are also currently about 4 x 10/8 meters apart.

The gravitational constant, G, is 6.67 x 10-/11 as always.

Question 1
If you were to weight yourself in the moon and the Earth, which would have the higher weight?

Physics
1 answer:
Basile [38]2 years ago
7 0

Answer

Gravity is what holds us down on the earth's (or moon's) surface. If you were to weigh yourself on a scale on Earth and then on the moon, the weight read on the moon would be 1/6 your earth weight

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People in the future may well live inside a rotating space structure that is more than 2 km in diameter. Inside the structure, p
Sergeeva-Olga [200]

Answer:

option B

Explanation:

given,

diameter of the rotating space = 2 Km

Force exerted at the edge of the space = 1 g

force experienced at the half way = ?

As the object is rotating in the circular part

Force is equal to centripetal acceleration.

at the edge

g = ω² r

ω is the angular velocity of the particle

r is the radius.

now, acceleration at the half way

g' = ω² r'

g' = \omega^2 (\dfrac{r}{2})

g' =\dfrac{1}{2}(\omega^2 r)

g'=\dfrac{g}{2}

People at the halfway experience g/2

hence, the correct answer is option B

7 0
3 years ago
A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of
Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

  1.29 kg/m³   =  <u>      mass    </u>

                             1800 m³

mass =  1.29 kg/m³  ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

8 0
3 years ago
Read 2 more answers
A4 kg bowling ball begins rolling down a at bowling alloy at 6 m/s . When it strikes the pins, it is estimated to be moving at 5
Paul [167]

Answer:

Energy lost due to friction is 22 J      

Explanation:

Mass of the ball m = 4 kg

Initially velocity of ball v = 6 m/sec

So kinetic energy of the ball KE=\frac{1}{2}mv^2

KE=\frac{1}{2}\times 4\times 6^2=72J

Now due to friction velocity decreases to 5 m/sec

Kinetic energy become

KE=\frac{1}{2}\times 4\times 5^2=50J

Therefore energy lost due to friction = 72 -50 = 22 J

8 0
3 years ago
Conductors have ___<br> resistance.
vazorg [7]

Answer:

little/no

Explanation:

Conductors are materials, which conduct electricity and/or heat. That means, that their resistance to such energy is so little, that an electric current is able to pass through.

5 0
3 years ago
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t
kozerog [31]

Answer:

a) 6 times farther.  b) 6 times longer.

Explanation:

Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:

vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s

In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:

vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s

Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:

y = voy*t - (\frac{1}{2}*\frac{g}{6} * t^{2} )

If the total displacement in the vertical direction is 0 (which means  that the time if the total time of flight), we can solve for t, as follows:

t = \frac{voy*12}{g} = \frac{9.72 m/s*12}{9.8m/s2} = 11. 9 s

On earth, this time could be calculated in the same way:

t = \frac{voy*12}{g} = \frac{9.72 m/s*2}{9.8m/s2} = 1.98 s

As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:

Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m

On earth, the distance travelled had been as follows:

Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m

⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00

b) As we have just found, the time of flight on the moon and on the earth are as follows:

tmoon = 11. 9 s

tearth = 1.98 s

⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0

8 0
2 years ago
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