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Digiron [165]
2 years ago
12

Planets A and B have the same size, mass, and direction of travel, but planet A is traveling through space at half the speed of

planet B. Which statement correctly explains the weight you would experience on each planet? A. You would weigh less on planet B because it is traveling twice as fast as planet A. B. You would weigh the same on both planets because their masses and the distance to their centers of gravity are the same. O C. You would weigh more on planet B because it is traveling twice as fast as planet A. D. You would weigh the same on both planets because your mass would adjust depending on the planet's speed.​
Physics
1 answer:
Ganezh [65]2 years ago
4 0

Answer:

B. You would weigh the same on both planets because their masses and the distance to their centers of gravity are the same.

Explanation:

Given that Planets A and B have the same size, mass.

Let the masses of the planets A and B are m_A and m_B respectively.

As masses are equal, so m_A=m_B\cdots(i).

Similarly, let the radii of the planets A and B are r_A and r_B respectively.

As radii are equal, so r_A=r_B\cdots(ii).

Let my mass is m.

As the weight of any object on the planet is equal to the gravitational force exerted by the planet on the object.

So, my weight on planet A, w_A= \frac {Gm_Am}{r_A^2}

my weight of planet B, w_B=\frac {Gm_Bm}{r_B^2}

By using equations (i) and (ii),

w_B=\frac {Gm_Am}{r_A^2}=w_A.

So, the weight on both planets is the same because their masses and the distance to their centers of gravity are the same.

Hence, option (B) is correct.

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Do you think there is water on exoplanets
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A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi
malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

1)

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude mg, where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

2)

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

F_{net}=ma

where

F_{net} is the net force

m = 75.0 kg is the mass of the bicyclist

a=2.20 m/s^2 is its acceleration

Solving, we find the net force:

F_{net}=(75.0)(2.20)=165 N

3)

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

F_{net}=F-R

where:

F is the forward force of push

R is the air drag

We know that:

F_{net}=165 N is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

F=F_{net}+R=165+60=225 N

6 0
3 years ago
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