<span>
No. of e- at a instant through given area is = total charge passing per
unit time/charge of e-
using i=q/t=(no.of e-(n))(1.6*10^-19)= 10
n=6.3*10^19 </span>
The gravitational force of the shell exerts is 4.25m x 10¯¹² N.
We need to know about gravitational force to solve this problem. The gravitational force is the force caused by two masses of objects. The magnitude of gravitational force can be determined as
F = G.m1.m2 / R²
where F is the gravitational force, G is the gravitational constant (6.674 × 10¯¹¹ Nm²/kg²), m1 and m2 are the mass of the object and R is the radius.
From the question above, we know that
m1 = 1.6 kg
m2 = m
R = 5.01 m
By substituting the following parameters, we get
F = G.m1.m2 / R²
F = 6.674 × 10¯¹¹ . 1.6 . m / 5.01²
F = 4.25m x 10¯¹² N
where m is the mass of the shell
For more on gravitational force at: brainly.com/question/19050897
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Answer:
Backwards
Explanation:
Friction always acts in the direction opposing motion so if the motion of the wagon is forward, friction acts backwards.
The more arms it has the less of a chance the prey has to swim away.
Answer:
The minimum possible coefficient of static friction between the tires and the ground is 0.64.
Explanation:
if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :
Fc = f
m×(v^2)/(R) = μ×m×g
(v^2)/(R) = g×μ
μ = (v^2)/(R×g)
= ((25)^2)/((100)×(9.8))
= 0.64
Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.
