Hello!
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?
Data:
For a spring (or an elastic), the elastic potential energy is calculated by the following expression:
Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.
Solving:
Answer:
The displacement of the spring = 0.8 m
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weight less on moon than on earth.
high on lift off - G force
low in orbit.
zero at a point between earth and moon
Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?
Answer:
17.32 m
Explanation:
From coulomb's Law,
F = kqq'/r²........................... Equation 1
Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.
make r the subject of the equation above
r = √(kqq'/F)..................... Equation 2
From the question,
Given: q = q' = 0.005 C, F = 750 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute these values into equation 2
r = √(9.0×10⁹×0.005×0.005/750)
r = √(300)
r = 17.32 m.
Hence the distance between the pair of charges = 17.32 m
1.b 2.c 3.a 4.b 5.d 6.a 7.a 9.idk 10.c hope thes helps have a nice day and God bless
