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alexandr402 [8]
3 years ago
14

What rate do things fall to Earth?

Physics
1 answer:
Dima020 [189]3 years ago
3 0

Answer:

9.8 meters per square second

Explanation:

Free Falling Object. the value of g is 9.8 meters per square second on the surface of the earth. The gravitational acceleration g decreases with the square of the distance from the center of the earth. But for many practical problems, we can assume this factor to be a constant.

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The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0
3 years ago
B) A satellite with mass m orbits the Earth at a radius r. A second satellite also with mass m orbits the
Snezhnost [94]

Answer:

So, given the eqn Fg=G(m1+m2/r^2) where G is the gravitational constant, m is the mass of the satellite and m2 is the mass of the earth and r is the distance from earth to the satellite, the force of earths gravity should be quartered.

Cause (2r)^2 gets turned into (4r^2) where 4r^2 is compared to r^2

Explanation:

6 0
3 years ago
The two properties of an electron that cannot be known exactly at the same time are the ____________________.
enyata [817]
Position and momentum.

This is Heisenberg's Uncertainty Principle:
Δx Δp ≥ h ÷ 4π,   where Δx is the change in position, Δp is the change in momentum, and h is Planck's Constant.
6 0
3 years ago
You need to fill a basketball with air to play with it. You know that your pump expels air at a velocity of 6 cm/s. You know tha
Scrat [10]

Answer:

The diameter of the needle is <u>4.675 cm</u>.

Explanation:

Given:

Volume flow rate is, \dot V=103\ cm^3

Velocity of air expelled by pump is, v=6\ cm/s

Let the area of the needle be 'A' cm² and the diameter be 'd' cm.

We know that, volume flow rate of the air expelled by pump is given as the product of the needle's area and velocity of air flowing through that area.

Therefore, volume flow rate is given as:

\dot V=A\times v\\\\103=A\times 6\\\\6A=103\\\\A=\frac{103}{6}\ cm^2

Now, considering the needle to be circular, area of the needle can be written as:

A=\frac{\pi}{4}\times d^2\\\\\frac{103}{6}=\frac{\pi}{4}\times d^2\\\\d^2=\frac{103\times 4}{6\pi}\\\\d=\sqrt{\frac{103\times 4}{6\pi}}\\\\d=4.675\ cm

Therefore, the diameter of the needle is 4.675 cm.

8 0
3 years ago
A mass on a string is swung in a circle of radius 0.75m at 7.0m/s.what its rate of acceleration.​
Ulleksa [173]

Explanation:

ac = v^2/r

= (7.0 m/s)^2/(0.75 m)

= 65 m/s^2

3 0
3 years ago
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