Hello!
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.
Data:
Hooke represented mathematically his theory with the equation:
F = K * Δx
On what:
F (elastic force) = 2 N
K (elastic constant) = 4 N/cm
Δx (deformation or elongation of the elastic medium or distance from a spring) = ?
Solving:
simplify by 2
Answer:
B.) 1/2 cm
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Answer:
Gravitational force mg' =(49)mg=49×63=28N.
Explanation:
What is the gravitational force on it due to the Earth , at a height equal to half the radius of the Earth ? (Given that the radius of Earth = 6400 km). Gravitational force mg ( This is not the real explanation)
Answer:
It can cause an object to accelerate.
It can cause an object to stop moving.
It can cause an object to start moving.
It can cause an object to change directions.
Explanation:
When the velocity of an object is increased in the same direction, the object is said to have positive acceleration. If it increases its velocity in a direction that is opposite to the original direction, it is negative acceleration.
When an object that's already moving is made to stop, it is said to have decelerated. Deceleration is negative acceleration.
When an object at rest is made to move by applying a force, it is said to have accelerated to some final velocity, during its motion for some duration.
An object at rest will remain at rest is said to have no net force acting on it.
Answer:
1.28 s
Explanation:
Given:
Δy = 8 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(8 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 1.28 s
