The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.
Now from Newton`s first law, the net force on gymnast,
![F_{net} =F-W=ma](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3DF-W%3Dma)
Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast (
acceleration due to gravity)
Therefore,
OR ![F=ma+mg=m(g+a)](https://tex.z-dn.net/?f=F%3Dma%2Bmg%3Dm%28g%2Ba%29)
Given
and![a=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}](https://tex.z-dn.net/?f=a%3D9%5Ctimes%20g%3D9%5Ctimes%209.8%20m%2Fs%5E%7B2%7D%20%3D88.2%20m%2Fs%5E%7B2%7D)
Substituting these values in above formula and calculate the force exerted by the gymnast,
![F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )](https://tex.z-dn.net/?f=F%3D%2840%20kg%29%20%2888.2%20m%2Fs%5E%7B2%7D%20%2B9.8%20m%2Fs%5E%7B2%7D%20%29)
![F=3.537\times10^{3}N](https://tex.z-dn.net/?f=F%3D3.537%5Ctimes10%5E%7B3%7DN)
It would be 17 m/s
If we use
V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.
Answer:
18.89cm
Explanation:
As we know that the person is standing 5m in front of the camera
![d_0=5m=500cm](https://tex.z-dn.net/?f=d_0%3D5m%3D500cm)
The focal length of the lens =50cm
f=50 cm
By Lens formula we have:
![\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%20%3D%20%5Cdfrac%7B1%7D%7Bd_i%7D%20%2B%20%5Cdfrac%7B1%7D%7Bd_o%7D%5C%5C%5Cdfrac%7B1%7D%7B50%7D%20%3D%20%5Cdfrac%7B1%7D%7Bd_i%7D%20%2B%20%5Cdfrac%7B1%7D%7B500%7D%5C%5C%5Cdfrac%7B1%7D%7Bd_i%7D%20%3D%5Cdfrac%7B1%7D%7B50%7D-%5Cdfrac%7B1%7D%7B500%7D%5C%5C%5Cdfrac%7B1%7D%7Bd_i%7D%3D0.018%5C%5Cd_i%3D55.56cm)
By the formula of magnification
![\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm](https://tex.z-dn.net/?f=%5Cdfrac%7Bh_i%7D%7Bh_o%7D%20%3D%20%5Cdfrac%7B55.56%7D%7B500%7D%5C%5C%5C%5Ch_i%20%3D%20%5Cdfrac%7B55.56%7D%7B500%7D%20%5Ctimes%20h_o%5C%5C%5C%5C%20h_o%3D1.70m%3D170cm%5C%5C%5C%5CTherefore%3A%20h_i%3D%5Cdfrac%7B55.56%7D%7B500%7D%20%5Ctimes%24%20170%20cm%5C%5C%5C%5Ch_i%20%3D18.89%20cm)
The height of the image formed is 18.89cm.
The tilt of the moon's axis does not allow for monthly alignment, so the lunar and solar eclipse do not happen every month.
<h3>How do the lunar and solar eclipse occur?</h3>
- For the occurrence of lunar and solar eclipse, the sun, moon and the earth must remain in a plan and along a straight line.
- When the earth appears in between the sun and the moon, lunar eclipse occurs.
- When the moon appears in between the sun and the earth, solar eclipse occurs.
- The moon and earth are rotating not only around the sun, but also around the black hole of Milky way galaxy.
- So they are not present in a plan as well as in a straight line in every full moon and new moon time.
Thus, we can conclude that the option D is correct.
Learn more about the lunar eclipse and solar eclipse here:
brainly.com/question/8643
#SPJ1