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Ann [662]
3 years ago
9

In what speed does an object fall, if after 10 seconds it hits the surface of Earth?​

Physics
1 answer:
Maru [420]3 years ago
8 0

Answer:

98m/s

Explanation:

Given parameters:

Time  = 10s

Unknown:

Final speed  = ?

Solution:

To solve this problem, we use the expression below;

          v  = u + gt

v is the final velocity

u is the initial velocity  = 0m/s

g is the acceleration due to gravity  = 9.8m/s²

t is the time

   so;

         v = 0 + 9.8 x 10  = 98m/s

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Objects with masses of 255 kg and a 555 kg are separated by 0.390 m.
eimsori [14]

Answer:

(a). The net gravitational force is 4.20\times10^{-6}\ N

(b). The position is at 0.232 m.

Explanation:

Given that,

Mass of one object M = 255 kg

Mass of another object M'= 555 kg

Separation = 0.390 m

(a). We need to calculate the net gravitational force

Using formula of force

F_{net}=\dfrac{GM'm}{r^2}-\dfrac{GMm}{r^2}

F_{net}=\dfrac{Gm(M'-M)}{r^2}

Put the value into the formula

F_{net}=\dfrac{6.67\times10^{-11}\times32.0(555-255)}{(0.390)^2}

F_{net}=4.20\times10^{-6}\ N

The net gravitational force is 4.20\times10^{-6}\ N

(b). We need to calculate the position

Force from 555 mass = Force from 255 mass

\dfrac{Gm\times555}{x^2}=\dfrac{Gm\times255}{(0.390-x)^2}

555(0.390-x)^2=255x^2

300x^2-0.78x+84.4155=0

x=0.232\ m

The position is at 0.232 m.

Hence, (a). The net gravitational force is 4.20\times10^{-6}\ N

(b). The position is at 0.232 m.

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