Answer:
50 N.
Explanation:
On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:
The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:
In conclusion, the normal force acting on the block is 50 N.
Answer:
The crest to trough distance = 8 m
Explanation:
Given that,
The amplitude of a particular wave is 4.0 m.
We need to find the crest to trough distance.
We know that,
Amplitude = The distance from the base line to the crest or the the distance from the baseline to the trough.
It means,
Distance from crest to trough = 2(Amplitude)
= 2(4)
= 8 m
Hence, the crest to trough distance is equal to 8 m.
Answer:
20.0 cm
Explanation:
Here is the complete question
The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Solution
Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.
Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.
Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m
Now, P' = 1/u + 1/v
1/u = P'- 1/v
1/u = 55.0 D - 1/0.02 m
1/u = 55.0 m⁻¹ - 1/0.02 m
1/u = 55.0 m⁻¹ - 50.0 m⁻¹
1/u = 5.0 m⁻¹
u = 1/5.0 m⁻¹
u = 0.2 m
u = 20 cm
So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.
