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12345 [234]
4 years ago
8

A bowling ball and a bag of potato chips are dropped at the same time from a tall tower on the moon. there is no air on the moon

. what happens?
Physics
2 answers:
rusak2 [61]4 years ago
8 0
Both objects will hit the ground at the same time
masya89 [10]4 years ago
3 0

Answer:

they both will hit the surface of moon at same time

Explanation:

When a bowling ball and potato chips are dropped at the same time on surface of moon

then during the motion above the surface of moon they both will have only force of gravity due to moon

so we have

F = mg_{moon}

now the we can say that the acceleration of object on this surface is given as

a = \frac{F}{m}

a = \frac{mg_{moon}}{m}

a = g_{moon}

since this acceleration on both objects is independent of their mass

so we can say that they both will hit the surface of moon at same time

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Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive
Burka [1]

Answer:

1.91773\times 10^{37}\ kg

Explanation:

v = Orbital speed = 130 km/s

d = Diameter = 16 ly

r = Radius = \dfrac{d}{2}=\dfrac{16}{2}=8\ ly

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

1\ ly=9.461\times 10^{15}\ m

As the centripetal force balances the gravitational energy we have the following relation

\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow M=\dfrac{v^2r}{G}\\\Rightarrow M=\dfrac{130000^2\times 8\times 9.461\times 10^{15}}{6.67\times 10^{-11}}\\\Rightarrow M=1.91773\times 10^{37}\ kg

Mass of the the massive object at the center of the Milky Way galaxy is 1.91773\times 10^{37}\ kg

4 0
3 years ago
Explain how the pressure at the bottom of a container depends on the container shape and the fluid height
3241004551 [841]

The pressure at the bottom of a column of fluid in a container
depends only on the depth of the fluid, not on the shape of the
container.  The pressure is simply the result of the weight of the
fluid resting on the bottom.

3 0
3 years ago
The first element on the periodic table is<br> Answer here
Gwar [14]

Answer:

Hydrogen

Explanation:

8 0
3 years ago
Read 2 more answers
A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
valentina_108 [34]

Answer:

E(4 \pi r^{2})={\frac{ Q r^{3}}{R^{3}\epsilon _{0}}}

or

E(4 \pi)={\frac{ Q r}{R^{3}\epsilon _{0}}}

Explanation:

We know that Gauss's law states that the Flux enclosed by a Gaussian surface is given by

E.S=\frac{q}{\epsilon_{0}}

Here , E is electric field and S is surface are and q is charge enclosed by the surface and e is electrical permeability of the medium.

Here the Gaussian is of radius r<R so area of surface is

S=4 \pi r^{2}

Also, charge enclosed by the surface is

Charge =\frac{Total \: Charge }{Total \:Volume} \times Volume \: of \: Gaussian \: surface

therefore,

q=\frac{Q}{\frac{4}{3} \pi R^{3} }\frac{4}{3} \pi r^{3} =\frac{ Q r^{3}}{R^{3}}

Here Q is total charge,

Insert values in Gauss's law

E(4 \pi r^{2})=\frac{\frac{ Q r^{3}}{R^{3}}}{\epsilon _{0}}

Rearrange them

E(4 \pi r^{2})={\frac{ Q r^{3}}{R^{3}\epsilon _{0}}}

on further solving

E(4 \pi)={\frac{ Q r}{R^{3}\epsilon _{0}}}

This is the required form.

6 0
4 years ago
An amusement park designer wants his swing to have a centripetal
Valentin [98]

Answer:

15.2 m

Explanation:

a = v² / r

19 m/s² = (17 m/s)² / r

r = 15.2 m

4 0
3 years ago
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