Answer:
The girl will move with constant velocity
Explanation:
If after a certain time t_0 the velocity of the girl is v_0 =gt_0 and the upward force on the girl due to rope is mg ,where g is gravitational acceleration. Then the girl will move down with the constant velocity v_0 .
The girl will move with constant velocity,as explained above.
The final speed of the combined masses is 0.21 m/s
Applying the law of conservation of momentum:
Total momentum before collision = Total momentum after collision.
⇒ Formula:
MU+mu = V(M+m).................. Equation 1
⇒ Where:
- M = mass of the first body
- m = mass of the second body
- U = Initial speed of the first body
- u = Initial speed of the second body
- V = common final speed.
From the question,
⇒ Given:
- M = 0.24 kg
- U = 0.60 m/s
- m = 0.26 kg
- u = -0.20 m/s (traveling in opposite direction)
⇒ Substitute these values into equation 1
- 0.24(0.6)+0.26(-0.20) = V(0.24+0.2)
⇒ Solve for V
- 0.144-0.052 = 0.44V
- 0.44V = 0.092
- V = 0.092/0.44
- V = 0.209
- V ≈ 0.21 m/s
Hence the final speed of the combined masses is 0.21 m/s
Learn more about speed here: brainly.com/question/4931057
Plasma consists of ionized particles I think
Explanation:
Given that,
Work done, W = 4.57 J
Distance, ![x_1=9.11\ cm](https://tex.z-dn.net/?f=x_1%3D9.11%5C%20cm)
Distance, ![x_2=9.54\ cm](https://tex.z-dn.net/?f=x_2%3D9.54%5C%20cm)
We know that the work done by the spring is given by :
![W_=-kx_1^2](https://tex.z-dn.net/?f=W_%3D-kx_1%5E2)
![k=\dfrac{W}{x_1^2}](https://tex.z-dn.net/?f=k%3D%5Cdfrac%7BW%7D%7Bx_1%5E2%7D)
![k=\dfrac{4.57}{(9.11\times 10^{-2})^2}](https://tex.z-dn.net/?f=k%3D%5Cdfrac%7B4.57%7D%7B%289.11%5Ctimes%2010%5E%7B-2%7D%29%5E2%7D)
k = 550.65 N/m
To find,
Let
is the extra work is required to stretch it an additional 9.54 cm. It can be calculated as :
![W_2=\dfrac{1}{2}k(x_2^2-x_1^2)](https://tex.z-dn.net/?f=W_2%3D%5Cdfrac%7B1%7D%7B2%7Dk%28x_2%5E2-x_1%5E2%29)
![W_2=\dfrac{1}{2}\times 550.65(9.54^2-9.11^2)](https://tex.z-dn.net/?f=W_2%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20550.65%289.54%5E2-9.11%5E2%29)
W = 2207.96 J
So, the extra wok done is 2207.96 J.