Jen falls out of a tree and lands on a trampoline. The trampoline sags 60 cm before launching Jen back into the air. At the very
1 answer:
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Answer:
7.55 km/s
Explanation:
The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:
where
is the gravitational constant
is the mass of the telescope
is the mass of the Earth
is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)
v = ? is the orbital velocity of the Hubble telescope
Re-arranging the equation and substituting numbers, we find the orbital velocity:
Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;
Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.
f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;
Here, p is the power of the lens.
Put f= 35.71 cm.
p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.