Can you sent a picture of the answers
Answer:
333.3 m
Explanation:
Given
![m =100g\ =\ 0.1kg\\v = 100 m/s\\g = 10 m/s ^2](https://tex.z-dn.net/?f=m%20%3D100g%5C%20%3D%5C%20%200.1kg%5C%5Cv%20%3D%20100%20m%2Fs%5C%5Cg%20%3D%2010%20m%2Fs%20%5E2)
Potential energy =
......Equation(1)
We know that
Potential energy=mgh
Kinetic energy =![\frac{1}{2} mv^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mv%5E%7B2%7D)
Now From the Equation(1)
![mgh=\frac{2}{3}*\frac{1}{2} mv^{2}\\ gh=\frac{v^{2} }{3} \\10 * h=\ \frac{10000}{3}\\ h=\ \frac{1000}{3} \\h=333.3\ m](https://tex.z-dn.net/?f=mgh%3D%5Cfrac%7B2%7D%7B3%7D%2A%5Cfrac%7B1%7D%7B2%7D%20mv%5E%7B2%7D%5C%5C%20%20gh%3D%5Cfrac%7Bv%5E%7B2%7D%20%7D%7B3%7D%20%5C%5C10%20%2A%20h%3D%5C%20%5Cfrac%7B10000%7D%7B3%7D%5C%5C%20h%3D%5C%20%5Cfrac%7B1000%7D%7B3%7D%20%5C%5Ch%3D333.3%5C%20%20m)
Answer:
d = 4 d₀o
Explanation:
We can solve this exercise using the relationship between work and the variation of kinetic energy
W = ΔK
In that case as the car stops v_f = 0
the work is
W = -fr d
we substitute
- fr d₀ = 0 - ½ m v₀²
d₀ = ½ m v₀² / fr
now they indicate that the vehicle is coming at twice the speed
v = 2 v₀
using the same expressions we find
d = ½ m (2v₀)² / fr
d = 4 (½ m v₀² / fr)
d = 4 d₀o
Momentum = mass x velocity
12 = 4 x v | ÷ both sides by 4
12 ÷ 4 =v
v= 3 m/s