Answer: 0.0267m/s
Explanation: Wave speed in m/s
Speed = Distance/Time
Substitute the given values into the formula
Speed = 0.1/3.75
Speed = 0.0267m/s
The answer is strong winds, i hoped this helped.
→if this helped please mark brainliest i need to level up←
I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
![F = G\frac{Mm}{r^{2} }](https://tex.z-dn.net/?f=F%20%3D%20G%5Cfrac%7BMm%7D%7Br%5E%7B2%7D%20%7D)
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ =
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ =
F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
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.
Answer:
![F_{braking}=337299 pdl](https://tex.z-dn.net/?f=F_%7Bbraking%7D%3D337299%20pdl)
Explanation:
Impulse-Momentum relation:
![I=\Delta p\\ F_{total}*t=m(v_{f}-v{o})](https://tex.z-dn.net/?f=I%3D%5CDelta%20p%5C%5C%20F_%7Btotal%7D%2At%3Dm%28v_%7Bf%7D-v%7Bo%7D%29)
![F_{total}=-F_{braking}+mgsin{\theta}](https://tex.z-dn.net/?f=F_%7Btotal%7D%3D-F_%7Bbraking%7D%2Bmgsin%7B%5Ctheta%7D)
We solve the equations in order to find the braking force:
![F_{braking}=m(v_{o}-v{f})/t+mgsin{\theta}=17926(75.6-30.3)/6.4+17926*32.17*sin21.4=337299 pdl](https://tex.z-dn.net/?f=F_%7Bbraking%7D%3Dm%28v_%7Bo%7D-v%7Bf%7D%29%2Ft%2Bmgsin%7B%5Ctheta%7D%3D17926%2875.6-30.3%29%2F6.4%2B17926%2A32.17%2Asin21.4%3D337299%20pdl)