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2 years ago

If you mass 35kg on earth what will your mass be on the moon where gravity is 1/6 that of earths. PLZZZZ HElP NEED ASAP

Physics

1 answer:

abruzzese [7]2 years ago

4 0

Answer:

Mass will be same on moon as on Earth but weight will be one-sixth of Earth.

Explanation:

Mass of a body doesn't depend on gravity. Mass is a constant quantity. So, mass on moon will be same as mass on Earth.

But, the weight of a body depends on gravity as weight is given as:

$\textrm{Weight}=\textrm{mass}\times \textrm{acceleration due to gravity}$

Therefore, if $g$ is acceleration due to gravity on Earth, then weight on Earth is, $W_{E}=mg$

Now, gravity on moon is one-sixth of Earth. So, $g_{moon}=\frac{1}{6}g$

Therefore, weight of the body on moon is, $W_{moon}=mg_{moon}=m\times \frac{1}{6}g=\frac{1}{6}mg=\frac{1}{6}W_{E}$

Therefore, a body has same mass both on moon and Earth but weight on moon is one-sixth of the weight on Earth.

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A satellite orbits Earth. The only force on the satellite is the gravitational force exerted by Earth. How does the satellite’s

bulgar [2K]

Answer:

here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

Explanation:

As we know that gravitational field is defined as the force experienced by the satellite per unit of mass

so we will have

$E = \frac{F}{m}$

now in order to find the acceleration of the satellite we know by Newton's II law

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so we will have

$a = \frac{F}{m}$

so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

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2 years ago

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2 years ago

The driver of a car slams on the brakes, causing the car to slow down at a rate of

sdas [7]

Answer:

A. The time taken for the car to stop is 3.14 secs

B. The initial velocity is 81.64 ft/s

Explanation:

Data obtained from the question include:

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Final velocity (V) = 0

Time (t) =?

Initial velocity (U) =?

A. Determination of the time taken for the car to stop.

Let us obtain an express for time (t)

Acceleration (a) = Velocity (V)/time(t)

a = V/t

Velocity (V) = distance (s) /time (t)

V = s/t

a = s/t^2

Cross multiply

a x t^2 = s

Divide both side by a

t^2 = s/a

Take the square root of both side

t = √(s/a)

Now we can obtain the time as follow

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Time (t) =..?

t = √(s/a)

t = √(256/26)

t = 3.14 secs

Therefore, the time taken for the car to stop is 3.14 secs

B. Determination of the initial speed of the car.

V = U + at

Final velocity (V) = 0

Deceleration (a) = –26ft/s2

Time (t) = 3.14 sec

Initial velocity (U) =.?

0 = U – 26x3.14

0 = U – 81.64

Collect like terms

U = 81.64 ft/s

Therefore, the initial velocity is 81.64 ft/s

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