Answer:
idk srry
Explanation:
i wish I could help you out
Answer:
v = 1.28 m/s
Explanation:
Given that,
Maximum compression of the spring, 
Spring constant, k = 800 N/m
Mass of the block, m = 0.2 kg
To find,
The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.
Solution,
When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,
Initial energy = Final energy

Substituting all the values in above equation,

v = 1.28 m/s
Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.
Answer:
the work required to turn the crank at the given revolutions is 8,483.4 J
Explanation:
Given;
torque required to turn the crank, T = 4.50 N.m
number of revolutions, = 300 turns
The work required to turn the crank is given as;
W = 2πT
W = 2 x 3.142 x 4.5
W = 28.278 J
1 revolution = 28.278 J
300 revlotions = ?
= 300 x 28.278 J
= 8,483.4 J
Therefore, the work required to turn the crank at the given revolutions is 8,483.4 J
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.