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3 years ago

How strongly the planet you're on pulls on you is your

2 answers:

6 0

The size of the forces between you and the planet you're on is
your weight on that planet.

Don't forget that you pull the planet with a force equal to the force
that the planet pulls on you. Your weight on Earth is the same as
the Earth's weight on you !

pogonyaev3 years ago

3 0

The answer to your question is well this is my guess, I think is gravity or weight

You might be interested in

A girl throws a tennis ball upward with an initial velocity of 4 m/s. What is the maximum displacement of the ball?

Jobisdone [24]

Answer:

0.82 m

Explanation:

The ball is in free fall - uniform accelerated motion with constant acceleration downward, $a=g=-9.8 m/s^2$ (acceleration of gravity). So we can use the following suvat equation to solve the problem:

$v^2-u^2=2as$

where

v is the final velocity

u = 4 m/s is the initial velocity

a is the acceleration

s is the displacement

At the maximum displacement, v = 0 (the velocity becomes zero). Substituting and solving for s, we find:

$s=-\frac{u^2}{2a}=-\frac{4^2}{2(-9.8)}=0.82 m$

8 0

3 years ago

A person supporting Eugene Talmadge for governor would have been MOST likely to belong to which occupation?

lisov135 [29]

Answer:

B) Farmer

Explanation:

tee hee i just got this question :)))

7 0

3 years ago

Read 2 more answers

a spiral spring of natural length 10.0cm has a scale pan hanging freely in its tower ends . When an object of mass of 20g is pla

weqwewe [10]

Answer:

Explanation:

There seems to be a typo in the problem statement. It says the spring stretches to a shorter length after more mass is added. Please check the problem statement. I'm going to do the calculations assuming that the first length should be 11.80 cm and the second length should be 12.05 cm.

Hooke's law states that the force needed to compress or extend a linear spring is:

F = kΔx, where k is the stiffness and Δx is the displacement.

When a 20g object is placed in the pan, the spring stretches to a length of 11.80 cm. The force of the spring is counteracting the weight of both the pan and the object. Therefore:

(m + 0.020) g = k (0.1180 - 0.100)

And when another 30g object is placed in the pan, the spring stretches to a length of 12.05 cm.

(m + 0.020 + 0.030) g = k (0.1205 - 0.100)

We now have two equations and two variables. If we divide the second equation by the first equation:

(m + 0.050) / (m + 0.020) = (0.1205 - 0.100) / (0.1180 - 0.100)

(m + 0.050) / (m + 0.020) = 0.02050 / 0.0180

0.0180 (m + 0.050) = 0.02050 (m + 0.020)

0.0180 m + 0.0009 = 0.02050 m + 0.00041

0.00049 = 0.0025 m

m = 0.196

The pan has a mass of 0.196 kg, or 196 g.

4 0

3 years ago

A 975-kg elevator accelerates upward at 0.754 m/s2, pulled by a cable of negligible mass. Find the tension force in the cable.

Zina [86]

To solve this problem we will apply the concepts of equilibrium and Newton's second law.

According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So

$\sum F = ma$