Answer:
The answer is choice A.
Explanation:
Assuming you are in a situation with a gravitational field. You can divide the motion of the bullet into two components. One horizontal and the other in the vertical.
-- the big flash of light and heat coming out of the head
of a match when it gets hot enough;
-- the explosion of a tiny bit of gunpowder that can send
a bullet many miles;
-- the energy captured from a few drops of burning gasoline
that moves a car;
-- the energy in the carbohydrates you eat that is used
to move you around;
This would be a cryogenic intermodal tank. These are used to store and transport HAZMAT gases that require storage under specific pressure and temperature parameters. Cryogenic Intermodal tanks have pressure of 25 Psi or less.
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.
We have that the spring constant is mathematically given as

Generally, the equation for angular velocity is mathematically given by

Where
k=spring constant
And

Therefore

Hence giving spring constant k

Generally
Mass of earth 
Period for on complete resolution of Earth around the Sun


Therefore


In conclusion
The effective spring constant of this simple harmonic motion is

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