All categories

3 years ago

What would be the weight of the moon if it were resting on the surface of the earth

1 answer:

7 0

We need to be careful here.
The calculation of the gravitational force between two objects
refers to the distance between their centers.
The minimum possible distance between the Earth's and moon's
centers is the sum of their radii (radiuses).

Earth's radius . . . . . 6,360 km = 6.36 x 10⁶ meters
Moon's radius . . . . . 1,738 km = 1.738 x 10⁶ meters
Sum of their radii = 8.098 x 10⁶ meters

Also:
Earth's mass . . . . . 5.972 x 10²⁴ kg
Moon's mass . . . . . 7.348 x 10²² kg
<span>
and now we're ready to go !

   Gravitational force =

   G M₁ M₂ / R²

= (6.67 x 10⁻¹¹ N-m²/kg²)(</span><span>5.972 x 10²⁴ kg)(7.348 x 10²² kg)/</span>(8.098 x 10⁶ m)²

= (6.67 · 5.972 · 7.348 / 8.098²) · (10²³) Newtons

= (I get ...)    4.463 x 10²³ Newtons

That's almost exactly 10²³ pounds

   = 50,153,000,000,000,000,000 tons.

Those are big numbers.
All I can say is: I wouldn't exactly call that "resting" on the surface".

You might be interested in

The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizont

Elza [17]

Answer:

1923 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 65 Kg

Radius (r) = 2.5 m

Velocity (v) = 8.6 m/s

Centripetal force (F) =?

The centripetal force, F, can be obtained by using the following formula:

F = mv²/r

F = 65 × 8.6² / 2.5

F = 65 × 73.96 / 2.5

F = 4807.4 / 2.5

F = 1922.96 ≈ 1923 N

Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N

3 0

3 years ago

A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upwar

sesenic [268]

Answer:

6.0 m below the top of the cliff

Explanation:

We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

$v^2-u^2 = 2gd$

where

u = 0 (it starts from rest)

g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)

h = 24 m is the distance covered

Solving for h,

$v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s$

So the ball thrown upward is launched with this initial velocity:

u = 21.7 m/s

From now on, we take instead upward as positive direction.

The vertical position of the ball dropped from the cliff at time t is

$y_1 = h - \frac{1}{2}gt^2$

While the vertical position of the ball thrown upward is

$y_2 = ut - \frac{1}{2}gt^2$

The two balls meet when

$y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s$

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

$y_1 = h -\frac{1}{2}gt^2 = 24-\frac{1}{2}(9.8)(1.11)^2=18.0 m$

So the distance below the top of the cliff is

$d=24.0 - 18.0 = 6.0 m$

4 0

3 years ago

A surveyor measures the distance across a river that flows straight north by the following method. Starting directly across from

bearhunter [10]

Answer:

The width of the river is $z = 60.62 \ m$

Explanation:

From the question we are told that

The distance of the base line is D = 130 m

The angle is $\theta = 25^o$

A diagram illustration the question is shown on the first uploaded image

Applying Trigonometric Rules for Right-angled Triangle,

$tan 25 = \frac{z}{130}$

Now making z the subject

$z = 130 * tan (25)$

$z = 60.62 \ m$

8 0

3 years ago

A 100 W engine runs for 1 minute. How much work does it do?

Svet_ta [14]

d 6000 j because every second multiplied by that would equal out to be 6000

4 0

3 years ago

Read 2 more answers

A 30.6 kilogram object is pulled by a horizontal force of 243 newtons causing it to accelerate at 5 meters per second squared. W

Citrus2011 [14]

Answer:

Ffriction = 90 N

coefficient = 0.3

Explanation:

First, note that the sum of all the forces in the x directions equals the mass multiplied by the acceleration in the x direction.

assuming the direction of the pulling force is positive,

243 N - Ffriction = m * a

m= 30.6 kg

a= 5 m/s/s

Ffriction= 243 - m*a

Ffriction= 243 - (30.6)(5)

Ffriction=90 N

The force of friction is equal to the coefficient of friction multiplied by the normal force on the object. Because the pulling force is completely horizontal, the normal force of the object is equal to its weight, which is m * g, or (30.6 kg)(9.8 m/s/s) = 299.88 N

Ffriction = coefficient * Fnormal

90 = coefficient * 299.88

coefficient = 0.3

8 0

2 years ago