Answer:
100Kg.m/s
Explanation:
From the question, we obtained the following information:
M= Mass = 25kg
V = Velocity = 4m/s
Momentum =?
Momentum = MV = 25x4= 100Kg.m/s
Answer:
After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
Explanation:
Area of a circle = πr²
where;
r is the circle radius
Differentiate the area with respect to time.

dr/dt = 4 ft/sec
after 12 seconds, the radius becomes = 
To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt


dA/dt = 1206.528 ft²/sec
Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
The spring constant is 4 N/m
Explanation:
When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:

where
F is the magnitude of the spring applied
k is the spring constant
x is the elongation of the spring, relative to its equilibrium position
For the spring in this problem, we have:
F = 0.12 N (force applied)
x = 3 cm = 0.03 m (elongation of the spring)
Therefore, we can solve the formula for k to find the spring constant:

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