A 45-kg box is being pushed a distance of 6.4 m across the floor by a force P with arrow whose magnitude is 170 N. The force P w
1 answer:
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Here we can use coulomb's law to find the force between two charges
As per coulombs law
]tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we have
now by using the above equation we have
so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.
Answer:
The spring constant = 104.82 N/m
The angular velocity of the bar when θ = 32° is 1.70 rad/s
Explanation:
From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:
Also;
Thus;
where;
= deflection in the spring
k = spring constant
b = remaining length in the rod
m = mass of the slender bar
g = acceleration due to gravity
Thus; the spring constant = 104.82 N/m
b
The angular velocity can be calculated by also using the conservation of energy;
Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s
