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Nuetrik [128]
3 years ago
11

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

finds that, under the same circumstances, the stone returns to his hand in 2T. In both cases, he throws the stone with the same initial velocity and it feels negligible air resistance. The acceleration due to gravity on planet X (in terms of g) is:_______
a) g/.
b) g/4.
c) 2g.
d) g/2.
e) g.
Physics
1 answer:
Pachacha [2.7K]3 years ago
5 0

Answer:

d) g/2

Explanation:

We need to use one of Newton's equations of motion to find the position of the stone at any time t.

x(t) = x₀(t) + ut - ¹/₂at²

Where

x₀(t) = initial position of the stone.

x(t) - x₀(t) = distance traveled by the stone at any time.

u = initial velocity of the stone

a = acceleration of the stone

t = time taken

On both planets, before the stone was thrown by the astronaut, x = 0 and t = 0.

=> 0 = x₀(t)

=> x₀(t) = 0

On earth, when the stone returns into the hand of the astronaut at time T on earth, x = 0.

=> 0 = 0 + uT - ¹/₂gT² (a = g)

=> uT = ¹/₂gT²

=> g = 2u/T

On planet X, when the stone returns into the hand of the astronaut, time = 2T , x = 0.

=> 0 = 0 + u(2T) - ¹/₂a(2T)²

=> 2uT = 2aT²

=> a = u/T

By comparing we see that a = g/2.

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Setler79 [48]
They will both hit the ground at the same time because gravitational acceleration for all objects is the same.
6 0
3 years ago
A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children
trasher [3.6K]

Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

                     

Explanation:

The centripetal acceleration is given by:

a_{c} = \frac{v^{2}}{r}

Where:

v^{2}: is the tangential speed = 9.50 m/s

r: is the distance = 6.00 m

Hence, the centripetal acceleration is:

a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}

Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

I hope it helps you!

3 0
2 years ago
Read 2 more answers
The gnaphosid spider Drassodes cupreus has evolved a pair of lensless eyes for detecting polarized light. Each eye is sensitive
BartSMP [9]

Answer:

Part A

The intensity is  I = 618 W/m^2  

Part B

The intensity is  I_1= 81.884 W/m^2

Explanation:

From the question we are told that

       The intensity of the light detected by first eye is I = 700 W/m^2

Now at initial state according the question the light  ray is perpendicular to the eye so it means that it is at 90° the eye

Now the first question is to obtain the intensity the first eye (the first in this case is the one focused on the light  )would detect when the head is rotated by 20°  its previous orientation

This  is mathematically evaluated  as

                   I = I_i cos^2 ( 20^o)

                    I = 700\  cos^2 (20)

                    I = 618 W/m^2  

Now the second  question is to obtain the intensity the first eye (the first eye  in this case is the one that is not  focused on  the light  )would detect when the head is rotated by 20°  its previous orientation

Now in this case the angle between the eye and the light is 90-20 = 70°

           So

               I_1 = 700 \  cos^2 (70)

                   I_1= 81.884 W/m^2

 

5 0
3 years ago
Please give answer with solution​
UNO [17]
250 m. for a longer explanation or solution look at this article, i’m sorry.
https://www.quora.com/A-projectile-is-thrown-so-it-travels-a-maximum-range-of-1000m-How-high-will-it-rise
3 0
2 years ago
For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same.
andrey2020 [161]

Answer:

he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Explanation:

In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values ​​where some natural frequency of the system coincides with the excitation frequency.

In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.

From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

7 0
3 years ago
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