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OLga [1]
2 years ago
9

. How many times higher could an astronaut jump on the Moon than on Earth if her takeoff speed is the same in both locations (gr

avitational acceleration on the Moon is about on-sixth of that on Earth)?
1) On the Moon, the jump height is about 3 times higher than on Earth.
2) On the Moon, the jump height is about 6 times higher than on Earth.
3) On the Moon, the jump height is about 36 times higher than on Earth.
4) On the Moon, the jump height is more than on Earth by about a times.
Physics
1 answer:
Helga [31]2 years ago
5 0

Answer:

On the Moon, the jump height is about 6 times higher than on Earth.

Explanation:

The maximum height reached by an object is given by the formula as follows :

H=\dfrac{v^2}{2g}

v is the speed of object

g is acceleration due to gravity on Earth's surface

If g' is acceleration on surface of Moon, g' = g/6

Let H' is the height reached by object on Moon, then,

H'=\dfrac{v^2}{2g'}\\\\H'=\dfrac{v^2}{2\times \dfrac{g}{6}}\\\\H'=6\times \dfrac{v^2}{2g}\\\\H'=6H

On the Moon, the jump height is about 6 times higher than on Earth. Hence, the correct option is (2).

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2 years ago
A net force of Fnet acts on an object, causing the object to accelerate at a rat of 9m/s/s. what is the acceleration rate in a s
Readme [11.4K]

The new acceleration is c) 18m/s/s

Explanation:

Net force, mass and acceleration of an object are related by Newton's second law of motion:

F=ma

where

F is the net force on the object

m is its mass

a is its acceleration

In the first trial of this problem, a net force of F_{net} is applied to the object, causing an acceleration of

a=9 m/s^2

Calling the mass of the object 'm', this means that

F_{net} = ma = 9m [N] (1)

In the second trial, the force applied is 2F_{net}, so we have

2F_{net} = ma'

where a' is the new acceleration. Substituting (1) into the second equation, we find:

2(9m) = ma' \rightarrow a' = 18 m/s^2

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Learn more about Newton's second law:

brainly.com/question/3820012

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7 0
3 years ago
Greg and Tabitha are jogging together. At time t = 0, they are both traveling at a speed of 4 m/s, but Greg is 8 m ahead of Tabi
Ray Of Light [21]

Answer:

The answer to question is below

Explanation:

Data

t1 = 0

vo = 4 m/s

a)

      d = vot + \frac{1}{2} at^{2}

      d = 4t + \frac{1}{2} (1)t^{2} + 8

      d = 8 + 4t + \frac{t^{2} }{2}

b)

      d = 4t + \frac{1}{2} (1.5)t^{2}

      d = 4t + 0.75t^{2}

c)    

       8 + 4t + \frac{t^{2} }{2}  =  4t + 0.75t^{2}

       4t + \frac{1}{2} t^{2} + 8 = 4t + \frac{1}{2} (1.5)t^{2} \\

       0.75t^{2} - 0.5t^{2}  = 8

       0.25t^{2} = 8

                      t² = 32

                      t = 5.66 s

d)

      d = 4(5.66) + 0.75(5.66)²

      d = 22.64 + 24.02

      d = 46.67 m

8 0
3 years ago
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