<h2>The distance between students is 2.46 m</h2>
Explanation:
The force of attraction due to Newton's gravitation law is
F = 
Here G is the gravitational constant
m₁ is the mass of one student
m₂ is the mass of second student .
and r is the distance between them
Thus r = 
If we substitute the values in the above equation
r = 
= 2.46 m
Answer:
the average force exerted by seatbelts on the passenger is 5625 N.
Explanation:
Given;
initial velocity of the car, u = 50 m/s
distance traveled by the car, s = 20 m
final velocity of the after coming to rest, v = 0
mass of the passenger, m = 90 kg
Determine the acceleration of the car as it hit the pile of dirt;
v² = u² + 2as
0 = 50² + (2 x 20)a
0 = 2500 + 40a
40a = -2500
a = -2500/40
a = -62.5 m/s²
The deceleration of the car is 62.5 m/s²
The force exerted on the passenger by the backward action of the car is calculated as follows;
F = ma
F = 90 x 62.5
F = 5625 N
Therefore, the average force exerted by seatbelts on the passenger is 5625 N.
Answer:
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Explanation:
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Answer:
Found in the nucleus, Has mass of one amu
