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Goryan [66]
3 years ago
15

The Fitness Gram push-up test is a measure of

Physics
1 answer:
Morgarella [4.7K]3 years ago
6 0

Answer:

The answer is B.

Explanation:

I meant B. not C so sorry

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NeTakaya
The answer is true...
4 0
3 years ago
A ball of mass m is dropped from rest from a height h and collides elastically with the floor, rebounding to its original height
Jet001 [13]

Answer:

Explanation:

Given

mass m collides elastically with floor and jump to its original height I .e.velocity of rebound is same as the initial velocity

Impulse imparted by steel ball to the ground is given by the change in momentum of steel ball

J=\Delta P

and impulse is product of average force and time

so we need time to calculate the average force. So, option e is correct

3 0
3 years ago
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The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




7 0
3 years ago
Read 2 more answers
How do scientists measure the amount of carbon dioxide in the atmosphere for dates prior to the 1955?
Debora [2.8K]

The scientists measured the amount of carbon dioxide in the atmosphere for dates prior to 1955, From air bubbles in glacial ice cores.

Almost 412 parts per million (ppm) of carbon dioxide are currently present in the Earth's atmosphere, and that number is increasing. This indicates an 11 per cent increase since 2000, when the concentration was close to 370 ppm, and a 47 per cent increase since the start of the Industrial Age when the concentration was close to 280 ppm. Most of the time, scientists do not consider Carbon dioxide as a percentage of the earth's atmosphere. It helps to picture a million equal pieces of an atmospheric gas sample.

Learn more about carbon dioxide here:

brainly.com/question/3049557

#SPJ4

6 0
1 year ago
yo this is due tomorrow help ya girl out : if a 2kg mass is accelerating at a rate of 2m/s^2 on a frictionless surface. how much
krok68 [10]
We Know, F = m.a
Here, m = 2 Kg
& a = 2 m/s²

Substitute for it, 
F = 2*2 Kgm/s²
F = 4 Kgm/s²
8 0
3 years ago
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