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ikadub [295]
2 years ago
14

A round asteroid has a surface gravity of .0288 m/s^2 if the mass of the asteroid is 1.10 x10^18 kg what is its radius

Physics
1 answer:
Oduvanchick [21]2 years ago
6 0

Answer:

r = 50.47 x 10³ m = 50.47 km

Explanation:

Using the formula for the acceleration due to gravity:

g = \frac{Gm}{r^2}

where,

g = acceleration due to gravity on the surface of asteroid = 0.0288 m/s²

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

m= mass of asteroid = 1.1 x 10¹⁸ kg

r = radius of asteroid = ?

Therefore,

0.0288\ m/s^2 = \frac{(6.67\ x\ 10^{-11}\ Nm^2/kg^2)(1.1\ x\ 10^{18}\ kg)}{r^2} \\\\r = \sqrt{\frac{(6.67\ x\ 10^{-11}\ Nm^2/kg^2)(1.1\ x\ 10^{18}\ kg)}{0.0288\ m/s^2}}

<u>r = 50.47 x 10³ m = 50.47 km</u>

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3 years ago
A 20-cm-long spring is attached to a wall. When pulled horizontally with a force of 100 N, the spring stretches to a length of 2
Delvig [45]

Answer:

the length of stretched spring  in cm is 22

Explanation:

given information:

spring length, x1 = 20 cm = 0.2 m

force, F = 100 N

the length of spring streches, x2 = 22 cm = 0.22 m

According to Hooke's law

F = - kΔx

k = F/*=(x2-x1)

  = 100/(0.22 - 0.20)

  = 5000 N/m

if the spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end

m = 10.2 kg

W = m g

    = 10.2 x 9.8

    = 99.96 N

F = - k Δx

Δx = F / k

     = 99.96 / 5000

     = 0.02

Δx = x2- x1

x2 = Δx + x1

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Work = Force times Distance
W = Fd

Given W = 750J, F = 125N;

750 = 125d

Solving for d:
d = 750/125
d = 6

The box moved a distance of 6 meters.
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