I think this is correct, but I am not entirely certain.
Find the force constant of the spring:
F = - KX
(0 - 62.4) = -K(0.172m)
-362.791 = -K
362.791 N/m = K
Find the work done in stretching the spring:
W = (1/2)KX
W = (1/2)(362.791)(0.172m)
W = 31.2 J
Answer:
a = 2m/s^2
Explanation:
Force (F) = 100 N
Mass (m) = 50 kg
Here,
F = m×a
100 = 50 × a
a = 100÷50
a = 2m/s^2
Thus, the acceleration on the cart is a = 2m/s^2
-TheUnknownScientist
A beat is an interference pattern between two sounds of slightly different frequencies, perceived as a periodic variation in volume whose rate is the difference of the two frequencies. Frequency beat is equal to,
The reference frequency in our case would be 392Hz, and since there is the possibility of the upper and lower range for the amount of beats per second that the two possible frequencies are heard would be
Therefore the two possible frequencies the piano wire is vibrating at, would be 396Hz and 388Hz
Answer:
Increases.
Explanation:
The electric potential increases when the two positive charges of same magnitude bring close to one charge to another because there is repulsive force between them due to same charge and when the two opposite charges move away from each other, the potential energy decreases. When two opposite charges are brought closer together, electric potential energy decreases while on the other hand, when we move opposite charges apart from each other than the work done against the attractive force that leads to an increase in electric potential energy.
