To solve this problem it is necessary to apply the concepts related to the Rotational Force described from the equilibrium and Newton's second law.
When there is equilibrium, the Force generated by the tension is equivalent to the Force of the Weight. However in rotation, the Weight must be equivalent to the Centrifugal Force and the tension, in other words:
Where
Angular velocity is equal to the Period, at this case Earth's period
Radius of the Earth
m = mass
= Force of Tension
Newton's second law
Replacing and re-arrange to find the Tension we have,
Therefore when Sneezy is on the equator he is in a circular orbit with a Force of tension of 503.26N
Answer:
The spring constant is 60,000 N
The total work done on it during the compression is 3 J
Explanation:
Given;
weight of the girl, W = 600 N
compression of the spring, x = 1 cm = 0.01 m
To determine the spring constant, we apply hook's law;
F = kx
where;
F is applied force or weight on the spring
k is the spring constant
x is the compression of the spring
k = F / x
k = 600 / 0.01
k = 60,000 N
The total work done on the spring = elastic potential energy of the spring, U;
U = ¹/₂kx²
U = ¹/₂(60000)(0.01)²
U = 3 J
Thus, the total work done on it during the compression is 3 J
Force = change in momentum / time, Force in opposite direction so negative
-1.41 x 10⁶ = (p₂ - 3.87 x 10⁷) / 9.55
p₂ = 2.52 x 10⁷ Ns
Answer:
The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2
Explanation:
Given:
F = 550 N
m = 1000 kg
To Find:
a = ?
Solution:
So by the equation by Newton's 2nd Law of Motion,
F = m x a
550 N = 1000 kg x a
a = 550 N/ 1000 kg
a = 0.55 m/s^2
Therefore,
The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2
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