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Fudgin [204]
2 years ago
6

An astronaut stands on the surface of an asteroid. The astronaut then jumps such that the astronaut is no longer in contact with

the surface. The astronaut falls back down to the surface after a short time interval. Which of the following forces CANNOT be neglected when analyzing the motion of the astronaut?

Physics
1 answer:
Anna71 [15]2 years ago
6 0

(D) The gravitational force between the astronaut and the asteroid.

Reason :

All the other forces given in the options, except (D), doesn't account for the motion of the astronaut. They are the forces that act between nucleons or atoms and neither of them accounts for an objects motion.

You might be interested in
The weight of an object is the same on two different planets. The mass of planet A is only sixty percent that of planet B. Find
natka813 [3]

Answer:

0.775

Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

F=G\frac{Mm}{R^2}

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

F_A=G\frac{M_Am}{R_A^2}

For planet B,

F_B=G\frac{M_Bm}{R_B^2}

We also know that the weight of the object on the two planets is the same, so

F_A = F_B

So we can write

G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}

We also know that the mass of planet A is only sixty percent that of planet B, so

M_A = 0.60 M_B

Substituting,

G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}

Now we can elimanate G, MB and m from the equation, and we get

\frac{0.60}{R_A^2}=\frac{1}{R_B^2}

So the ratio between the radii of the two planets is

\frac{R_A}{R_B}=\sqrt{0.60}=0.775

6 0
1 year ago
When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both startfrom rest at the bo
Natali5045456 [20]

Answer:

46.4 s

Explanation:

5 minutes = 60 * 5 = 300 seconds

Let g = 9.8 m/s2. And \theta be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim .  Both of their motions are subjected to parallel component of the gravitational acceleration gsin\theta

Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8

As both of them cover the same distance

45000a = 3at_J^2/8

t_J^2 = 45000*8/3 = 120000

t_J = \sqrt{120000} = 346.4 s

So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time

7 0
2 years ago
Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.
Rufina [12.5K]

Answer: 1. decreasing the mass of both objects

2. decreasing the mass of one of the objects

3. increasing the distance between the objects

Explanation: Hope that helped! (:

4 0
2 years ago
A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
2 years ago
50 J of work was performed in 20 seconds. How much power was used to do this task?
yuradex [85]
Power=work/time
power=50/20
50/20=2.5
Therefore A. 2.5 W
7 0
1 year ago
Read 2 more answers
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