When you stretch a spring 13 cm past its natural length, it exerts a force of 21 N. What is the spring constant of this spring?
2 answers:
Answer:
Explanation:
From Hooke's law,
Substistuting the values, you can ignore the minus sign, since it is there to state that the force is always in the opposite direction of the deformation, and has no influence otherwise, and your constant is
Answer:
Spring constant, k = 161.5 N-m
Explanation:
It is given that,
Force on spring, F = 21 N
The spring is stretched 13 cm past its natural length i.e. x = 0.13 m
The Hooke's law gives the relationship between the force and the compression in the spring. Mathematically, it can be written as :
k is spring constant
-ve sign shows the direction of force is opposite
So,
k = 161.5 N-m
Hence, this is the required solution.
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Answer:
The acceleration of the proton is 9.353 x 10⁸ m/s²
Explanation:
Given;
speed of the proton, u = 6.5 m/s
magnetic field strength, B = 1.5 T
The force of the proton is given by;
F = ma = qvB(sin90°)
ma = qvB
where;
m is mass of the proton, = 1.67 x 10⁻²⁷ kg
charge of the proton, q = 1.602 x 10⁻¹⁹ C
The acceleration of the proton is given by;
Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²