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PilotLPTM [1.2K]
2 years ago
5

What is the weight of an astronaut with a mass of 75 kg on the moon? The gravitational field strength on the moon is 1.6 Nkg-1.

Physics
1 answer:
bearhunter [10]2 years ago
7 0

okay this is kinda easy

<u>What is the gravitational field strength on the moon?</u>

The Moon has a gravitational field strength of 1.6 N/kg.

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notsponge [240]

Answer:

yeah i knowwwwwwwwwwwwww

Explanation:

7 0
2 years ago
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Rearrange the momentum equation to solve for speed. Show all work.
nasty-shy [4]

Definition:                   Momentum  =  (mass) x (speed)


OK. Here we go.
Watch closely:


Divide each side        
by  'mass' :                  <span>Momentum / mass  =  Speed </span>


Did you follow that ?

3 0
3 years ago
A baseball player slides into third base with an initial speed of 4.0 m/s. If the coefficient of
musickatia [10]
The frictional force is given by F = μmg 

<span>where μ is the coeficient of friction. </span>

<span>Work done by frictional force = Fd = μmgd </span>

<span>Kinetic energy "lost" = 1/2 mv² </span>


<span>Fd = μmgd = 1/2 mv² </span>

<span>The m's cancel μgd = v² / 2 </span>



<span>d = v² / 2μg </span>

<span>d = 8² / 2(0.41)(9.8) </span>

<span>d = 32 / (0.41)(9.8) </span>

<span>d = 7.96 </span>

<span>Player slides 8 m . </span>



<span>Note. In your other example μ = 0.46 and v = 4 m/s </span>



<span>d = v² / 2μg </span>




<span>= 4² / 2(0.46)(9.8) </span>

<span>= 8 / (0.46)(9.8) </span>

<span>= 1.77 or 1.8 m.
</span>
Hope i Helped :D
3 0
3 years ago
Read 2 more answers
A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

\theta = \omega t\\\\\theta = (100 \frac{rev}{\min}  \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi  \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

6 0
3 years ago
a lawn mower is pushed with a force of 50n. if the angle between the handle of the mower and the ground is 30°. why doesn't the
Mademuasel [1]

Answer:

IT doesnt push down because the lawnmower has wheels and the ground is hard

Explanation:

Branliest

5 0
3 years ago
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