true if you are refering to the desing of the experimnt as it does identify the variable
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:
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.
It depends most on the temperature of the gas.
An object in motion will continue to move in the same direction and with the same speed unless acted upon by an unbalanced force. states that forces occur as equal and opposite pairs. The strength of the force is related to the mass of the objects and the distance between them.
