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worty [1.4K]
3 years ago
13

The gravitational acceleration on the moon is about one-sixth the size of the gravitational acceleration on Earth. According to

Newton’s second law of motion, what happens to an astronaut who goes to the moon?
Physics
2 answers:
LiRa [457]3 years ago
7 0

Answer:

the astronaut will experience his weight 1/6 times smaller than the weigh near the surface of earth

Explanation:

As per Newton's II law we know that net force on an object is given by the product of mass and acceleration

so here we have

F_{net} = ma

now we know that acceleration due to gravity on the surface of the moon is 1/6 times the acceleration near the surface of earth

so we will have

a = \frac{g}{6}

now the force near the surface of moon on an astronaut is given as

F = m\frac{g}{6}

so the astronaut will experience his weight 1/6 times smaller than the weigh near the surface of earth

Rainbow [258]3 years ago
5 0
<span>His weight decreases because the moon’s gravitational acceleration is less.</span>
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A Web site that ends in .gov or .edu is probably unreliable.<br> True<br> False
Aleonysh [2.5K]

Answer:

false, they are more reliable.

Explanation:

i had to do it last year.

8 0
3 years ago
Read 2 more answers
By what factor does the peak frequency change if the celsius temperature of an object is doubled from 20.0 ∘c to 40.0 ∘c?
mart [117]

Answer:

it increases by a factor 1.07

Explanation:

The peak wavelength of an object is given by Wien's displacement law:

\lambda=\frac{b}{T} (1)

where

b is the Wien's displacement constant

T is the temperature (in Kelvins) of the object

given the relationship between frequency and wavelength of an electromagnetic wave:

f=\frac{c}{\lambda}

where c is the speed of light, we can rewrite (1) as

\frac{c}{f}=\frac{b}{T}\\f=\frac{Tc}{b}

So the peak frequency is directly proportional to the temperature in Kelvin.

In this problem, the temperature of the object changes from

T_1 = 20.0^{\circ}+273=293 K

to

T_2 = 40.0^{\circ}+273 = 313 K

so the peak frequency changes by a factor

\frac{f_2}{f_1} \propto \frac{T_2}{T_1}=\frac{313 K}{293 K}=1.07

8 0
3 years ago
A mass is oscillating horizontally on a spring. At the locations A, B, C, D, and E, photogates are used to measure the speed of
svetoff [14.1K]

Complete Question

The complete question is is shown on the first uploaded

Answer:

The elastic potential energy at point B is  PE_{elastic} = 50J

The kinetic energy at point D is KE = 75J

Explanation:

Looking at the given point we can observe that mechanically energy(i.e potential and kinetic energy ) is conserved and it value is E_ {m} = 100J

     So at point B

           E_{m} = PE_{elastic} +KE

           100 = PE_{elastic} + KE

   KE at point B is  50J

So     PE_{elastic} = 100 - 50 = 50J

     Now at point D

          E_{m} = PE_{elastic} +KE

    PE_{elastic} at point D is 25J

 So  KE = 100 - 25 = 75J

4 0
3 years ago
At what height does a 1000-kg mass have potential energy of 1J relative to the ground?
Dvinal [7]

Answer:

0.0001 m

Explanation:

Given

M=1000 KG

PE=1 J

G=9.8 m/s

H=?

Formula

PE = MGH

H= PE/MG

H=1/1000x9.8

H= 1/9800

or 0.00010204

7 0
3 years ago
If the maximum length is 0.3 meters from the equillibrium point what is the and the spring oscillates 100 per minute what is it'
kaheart [24]
20.......................................
5 0
4 years ago
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