Explanation:
We have,
Mass of an object is 0.5 kg
Force constant of the spring is 157 N/m
The object is released from rest when the spring is compressed 0.19 m.
(A) The force acting on the object is given by :
F = kx
(B) The force is simply given by :
F = ma
a is acceleration at that instant
It is the mathematical and conceptual framework for contemporary Elementary practical physicas. if that make sense to you.
The velocity is 14 m/s
The parameters given on the question are
mass= 0.060 kg
kinetic energy= 5.9 joules
K.E= 1/2mv²
5.9= 1/2 × 0.060 × v²
5.9= 0.5 × 0.060v²
5.9= 003v²
v²= 5.9/0.03
v²= 196.66
v= √196.66
v= 14 m/s
Hence the velocity of the egg before it strikes the ground is 14 m/s
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Answer:
1.5X10^-4C
Explanation:
Expression for the electric force between the two charges is given by -
F = (k*q1*q2) / r^2
Here, k = constant = 9 x 10^9 N*m^2 / C^2
q1 = unknown
F=25N
q2 = 1.9x10^-6 C
r = 0.32 m
Substitute the given values in the above expression -
25=9x10^9 *(q1) * 1.9 x10^ -6/ (0.32m)^2
25= 17,100* (q1)/ 0.1024
multiply both sides by 0.1024 to get rid of the denominator
2.56=17,100*(q1)
Divide both sides by 17,100 to isolate q1
q1=1.5x10^-4C
