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hammer [34]
2 years ago
8

Health education provides

Physics
1 answer:
natka813 [3]2 years ago
5 0
I think the answer will be B
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Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
4 0
3 years ago
If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance, how
taurus [48]
T=2s
g=10m/s2
h=?
free fall: h=gt2/2
              h= 10*4/2
               h=40/2
              h=20m








3 0
3 years ago
What is the awnser to Blank occurs when a substance changes from a liquid to a gas
PolarNik [594]
Evaporation (or another word to use is water vapor.)
3 0
3 years ago
PLEASE HELP!!!!!
never [62]

Answer:

The answer is A

Explanation:

When a rockets thrusters push on the ground the ground pushes back on the rocket with equal force in the opposite direction. Hence the rocket takes off.

Newtons third law of motion states, for every action there is an equal and opposite reaction.

7 0
2 years ago
An object is moving in a straight line along the y axis. As a function of time, its position is given by the equation y=3.0+4.8x
romanna [79]

Answer:

Explanation:

<u>Instant Velocity and Acceleration </u>

Give the position of an object as a function of time y(x), the instant velocity can be obtained by

v(x)=y'(x)

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

a(x)=v'(x)=y''(x)

We are given the function for y

y(x)=3.0+4.8x +6.4x^2

Note we have changed the last term to be quadratic, so the question has more sense.

The velocity is

v(x)=y'(x)=4.8+12.8x

And the acceleration is a(x)=v'(x)=12.8

5 0
3 years ago
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