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olga55 [171]
3 years ago
11

Which quantities below of a solid object on this planet are NOT the same as on Earth?

Physics
1 answer:
AysviL [449]3 years ago
8 0

Answer:

Weight, acceleration when it falls vertically, are not same as that of earth.

Explanation:

Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.

So, the weight of object is not same as that on earth.

The mass is defined as the amount of matter contained in the object.

So, the mass of the object is same as that of earth.  

The volume of the object is defined as the space occupied by the object.

So, the volume of the object is same as that of earth.  

The density is defined as the ratio of mass of the object to its volume.

So, the density of the object is same as that of earth.  

The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.

So, the acceleration is not same as that of earth.

The color of the object is its characteristic.

It is same as that of earth.

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An earthquake is a natural rapid shaking of the _______ caused by the release of stored energy in rocks
REY [17]

Answer:

An earthquake is a natural rapid shaking of the tectonic plates caused by the release of stored energy in rocks

Explanation:

there are tectonic plates and the earthquake is caused by that energy the plates are always moving so it might get stuck in between each other and store the energy.

8 0
2 years ago
Why is the motion of an athlete moving along the circular path with Constant speed considered to be an accelerated motion?
Strike441 [17]

The speed is changing its direction all the time. There is an acceleration which changes the direction of the speed – that is called centripetal acceleration. Only uniform linear motions are considered to have no acceleration.

This is the general formula for acceleration

a = dv/dt

When calculating dv, you should keep in mind the change in the velocity vector’s direction. You can easily see in a graph that with dt tending to 0 (so the length of the arc covered is also tending to 0), the difference between vectors Vf and V0 has a direction which is perpendicular to velocity (the shorter the arc, the closest the angle is to 90 degrees).

There is a formula (which can be deducted from the previous formula) which allows you to calculate the acceleration:

a = v^2/r

Let’s talk about the units:

v is in m/s

r is in m

so v^2/r

is in (m/s)^2/m = (m^2/s^2)/m = m/s^2

which is the same unit as dv/dt:

dv/dt = (m/s)/s= m/s^2

5 0
3 years ago
A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r <
Svet_ta [14]

Answer:  

Hence the answer is E inside = KQr_{1} /R^{3}.

Explanation:  

E inside = KQr_{1} /R^{3}  

so if r1 will be the same then  

E  \begin{bmatrix}Blank Equation\end{bmatrix} proportional to 1/R3  

so if R become 2R  

E becomes 1/8 of the initial electric field.

8 0
3 years ago
Read 2 more answers
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

6 0
3 years ago
Read 2 more answers
PLS SOMEONE HELP QUICK
SVETLANKA909090 [29]

Answer:

the rates of rock formation are similar. i could be wrong tho.....

Explanation:

6 0
3 years ago
Read 2 more answers
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