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dimulka [17.4K]
2 years ago
6

A space probe lands on a newly discovered planet. A small canister is released from the probe and falls a distance of 3 m in 0.5

sec. What is the value for the acceleration of gravity on this new planet?
Physics
1 answer:
ivanzaharov [21]2 years ago
3 0

Answer:

24m/s²

Explanation:

Given

Distance S = 3m

Time of fall = 0.5sec

Required

Acceleration due to gravity

Using the equation of motion

S = ut+1/2gt²

Substitute the given values

3 = 0+1/2g(0.5)²

3 = 1/2(0.25)g

3 = 0.125g

g = 3/0.125

g = 24

Hence the value for the acceleration of gravity on this new planet is 24m/s²

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7 0
3 years ago
PLEASE ANSWER ASAP!!!!!!!
Viefleur [7K]

Answer:

B. Mechanical energy= 50J+30J=80J

4 0
3 years ago
I am struggling on this physics question. Brainly is my last hope. Could somebody please provide an answer to this question, wit
Aleks [24]

1) 29.4 N

The force of gravity between two objects is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M and m are the masses of the two objects

r is the separation between the centres of mass of the two objects

In this problem, we have

M=5.97\cdot 10^{24} kg (mass of the Earth)

m=3.0 kg (mass of the box)

r=R=6.37\cdot 10^6 m (Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)

Substituting into the equation, we find F:

F=\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24})(3.0)}{(6.37\cdot 10^6)^2}=29.4 N

2) g=9.8 m/s^2

Let's now calculate the ratio F/m. We have:

F = 29.4 N

m = 3.0 kg

Subsituting, we find

\frac{F}{m}=\frac{29.4}{3.0}=9.8 N/kg = 9.8 m/s^2

This is called acceleration of gravity, and it is the acceleration at which every object falls near the Earth's surface. It is indicated with the symbol g.

We can prove that this is the acceleration of the object: in fact, according to Newton's second law,

F=ma

where a is the acceleration of the object. Re-arranging,

a=\frac{F}{m}

which is exactly equal to the quantity we have calculated above.

5 0
3 years ago
A boxer punches a sheet of paper in mid air, and brings it from rest up to a speed of 25 m/s in 0.05 s. if the mass of the paper
zepelin [54]
Ok, so you've got to figure out a force F and you have the speed in which the boxer punches on determinate time and the mass of the sheet of paper.

So based on the formula that says that the Force is equal to the mass multiplied by the acceleration => F=ma.
You look at it and see that you only have mass which is measured on KG so there is no problem.
then you have the acceleration which is measured on meters and is defined by: a = Δv/Δt 
So now you can replace the velocity and the time you have there
⇒ a 25m/s / 0.05s 
you have computing that ⇒ 50m because the seconds were cancelled out.
and then you plug the meters into the force equation.
F=(0.005kg)(50)
F=0.25N
so the boxer will have a force of 0.25 Newton's.

6 0
3 years ago
Read 2 more answers
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