A block of ice of mass 4.30 kg is placed against a horizontal spring that has a force constant k = 250 N/m and is compressed a d
1 answer:
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Answer:
Electrical force, F = 90 N
Explanation:
It is given that,
Charge on sphere 1,
Charge on sphere 2,
Distance between two spheres, d = 6 cm = 0.06 m
Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,
F = 90 N
So, the electrical force between them is 90 N. Hence, this is the required solution.
Answer:
A) 31 kJ
B) 1.92 KJ
C) 40 , 2.48
Explanation:
weight of person ( m ) = 79 kg
height of jump ( h ) = 0.510 m
Compression of joint material ( d ) = 1.30 cm ≈ 0.013 m
A) calculate the force
Fd = mgh
F = mgh / d
W = mg
F(net) = W + F = mg ( 1 +
= 79 * 9.81 ( 1 + (0.51 / 0.013) )
= 774.99 ( 40.231 ) ≈ 31 KJ
B) calculate the force when the stopping distance = 0.345 m
d = 0.345 m
Fd = mgh hence F = mgh / d
F(net) = W + F = mg ( 1 +
= 79 * 9.81 ( 1 + (0.51 / 0.345) )
= 774.99 ( 2.478 ) = 1.92 KJ
C) Ratio of force in part a with weight of person
= 31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40
Ratio of force in part b with weight of person
= 1920 / 774.99 = 2.48