Answer:
50 N/m
Explanation:
Elastic energy = kinetic energy
EE = KE
½ kx² = ½ mv²
½ k (4 m)² = ½ (8.0 kg) (10.0 m/s)²
k = 50 N/m
Answer:
![F_2 = 1.10 \mu N](https://tex.z-dn.net/?f=F_2%20%3D%201.10%20%5Cmu%20N)
Explanation:
As we know that the electrostatic force is a based upon inverse square law
so we have
![F = \frac{kq_1q_2}{r^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7Br%5E2%7D)
now since it depends inverse on the square of the distance so we can say
![\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}](https://tex.z-dn.net/?f=%5Cfrac%7BF_1%7D%7BF_2%7D%20%3D%20%5Cfrac%7Br_2%5E2%7D%7Br_1%5E2%7D)
now we know that
![r_2 = 18.2 mm](https://tex.z-dn.net/?f=r_2%20%3D%2018.2%20mm)
![r_1 = 12.2 mm](https://tex.z-dn.net/?f=r_1%20%3D%2012.2%20mm)
also we know that
![F_1 = 2.45 \mu N](https://tex.z-dn.net/?f=F_1%20%3D%202.45%20%5Cmu%20N)
now from above equation we have
![F_2 = \frac{r_1^2}{r_2^2} F_1](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7Br_1%5E2%7D%7Br_2%5E2%7D%20F_1)
![F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7B12.2%5E2%7D%7B18.2%5E2%7D%282.45%5Cmu%20N%29)
![F_2 = 1.10 \mu N](https://tex.z-dn.net/?f=F_2%20%3D%201.10%20%5Cmu%20N)
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Answer:
<em>The change in momentum of the car is 3575 Kg.m/s</em>
Explanation:
<u>Impulse and Momentum</u>
The impulse (J) experienced by the object equals the change in momentum of the object (Δp).
The formula that represents the above statement is:
J = Δp
The impulse is calculated as
J = F.t
Where F is the applied force and t is the time.
The car hits a wall with a force of F=6500 N and stops in 0.55 s. Thus, the impulse is:
J = 6500 * 0.55
J = 3575 Kg.m/s
The change in momentum of the car is:
![\Delta p= J = 3575\ Kg.m/s](https://tex.z-dn.net/?f=%5CDelta%20p%3D%20J%20%3D%203575%5C%20Kg.m%2Fs)
The change in momentum of the car is 3575 Kg.m/s