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fiasKO [112]
3 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]3 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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7) T F If two forces of equal magnitude act on an object that is hinged at a pivot, the force acting farther from the pivot must
saul85 [17]

Answer:

False

Explanation:

The torque exerted by a force is given by:

\tau=Fd sin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the pivot

\theta is the angle between the directions of F and d

We see that the magnitude of the torque depends on 3 factors. In this problem, we have 2 forces of equal magnitude (so, equal F). Moreover, one of the forces (let's call it force 1) acts farther from the pivot than force 2, so we have

d_1 > d_2

However, this does not mean that force 1 produces a greater torque. In fact, it also depends on the angle at which the force is applied. For instance, if the first force is applied parallel to d, then we have

\theta_1 =0\\sin \theta=0

and the torque produced by this force would be zero.

So, the statement is false.

4 0
3 years ago
A 5.18x10^4 kg railroad car moves on a frictionless horizontal rails until it hits a horizontal spring stopper with a force cons
nalin [4]

Answer:

The initial velocity of the cart was 0.91m/s.

Explanation:

When the car comes to a stop it has delivered all of its kinetic energy to the spring; therefore,

\dfrac{1}{2}kx^2 = \dfrac{1}{2}  mv^2

solving for v we get:

v = x\sqrt{\dfrac{k}{m} },\\

we put in k = 4.21*10^{5}N/m, m = 5.18*10^{4}kg, and x = 0.32m to get:

v = (0.32m)\sqrt{\dfrac{4.21*10^5}{5.18*10^4} },\\

\boxed{v = 0.91m/s.}

4 0
4 years ago
A uniform stick has a mass of 4.42 kg and a length of 1.23 m. It is initially lying flat at rest on a frictionless horizontal su
nataly862011 [7]

10.7 rad/s is the final angular velocity of the stick.

Given:

Mass of the stick = 4.42 kg

Length of the stick = 1.23m

Force of impulse (I) = 12.8 N s

The linear velocity of the stick, v=\frac{I}{m}

                                                  v=\frac{12.8 N.s (\frac{1 kg m/s^2}{1 N}) }{4.42 kg}

                                                  v  = 2.89 m/s

Therefore, the final linear velocity of the stick is 2.89 m/s

∴w=\frac{12 Ir}{ml^{2} }

w=\frac{12 ( 12.8 N.s ) ( 46.4 cm)}{(4.42 kg) (1.23 m)^2}

w= \frac{12 (12.8 N.s) (46.4 cm) (\frac{10^-^2 m}{1 cm} )}{(4.42 kg) (1.23m)2}

w=10.7   rad/s

Therefore, 10.7 rad/s is the final angular velocity of the stick.

Learn more about linear velocity here:

brainly.com/question/15154527

#SPJ4

                                                 

5 0
2 years ago
Which situation is an an example of competition that could be found in the grassland
astraxan [27]
A good answer is a giraffe and a tree

4 0
4 years ago
Read 2 more answers
A joule, which is a unit of work, is equal to?
sweet-ann [11.9K]
Option A is correct
4 0
3 years ago
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