Answer:
-4.71 m/s
Explanation:
Given:
y₀ = 1.13 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2a (y − y₀)
v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.13 m)
v = -4.71 m/s
Answer:
They could be jumping off structures, running or climbing on rocks. They also could be participating in sports like soccer.
Explanation:
sources:
https://www.philippinesbasiceducation.us/2013/06/physical-activity-and-physical.html - about there play schedule
http://sportphil.com/category/articles/ - concil of philippines
Sorry that's all I could find.
Solution
In this question we have given
Charge,![q_{1}=-5\times 10^-9C](https://tex.z-dn.net/?f=q_%7B1%7D%3D-5%5Ctimes%2010%5E-9C)
Charge,![q_{2}=-1.5\times 10^-9C](https://tex.z-dn.net/?f=q_%7B2%7D%3D-1.5%5Ctimes%2010%5E-9C)
Charge,![q_{3} =1.1\times 10^-8C](https://tex.z-dn.net/?f=q_%7B3%7D%20%3D1.1%5Ctimes%2010%5E-8C)
Distance between Charge
and
,d=43cm=.43m
Let the distance between charge
and
, is x
Therefore,distance between charge
and
, will be=(.43-x)
We know By Couloms law, Force on Charge
due to charge
which are seperated by distance d is given by following equation
............(1)
Here, K=![9\times 10^9 Nm^2C^-2](https://tex.z-dn.net/?f=9%5Ctimes%2010%5E9%20Nm%5E2C%5E-2)
Therefore,
Force on Charge
due to charge ![q_{1}](https://tex.z-dn.net/?f=q_%7B1%7D)
..(2)
Similarly,
Force on Charge
due to charge ![q_{2}](https://tex.z-dn.net/?f=q_%7B2%7D)
..(2)
In equilibrium condition,
![F_{1} =F_{2}](https://tex.z-dn.net/?f=F_%7B1%7D%20%3DF_%7B2%7D)
![-\frac{k\times 5\times 10^-9C\times 1.1\times 10^-8C}{x^2}=-\frac{k\times 1.5\times 10^-9C\times 1.1\times 10^-8C}{(.43-x)^2}](https://tex.z-dn.net/?f=-%5Cfrac%7Bk%5Ctimes%205%5Ctimes%2010%5E-9C%5Ctimes%201.1%5Ctimes%2010%5E-8C%7D%7Bx%5E2%7D%3D-%5Cfrac%7Bk%5Ctimes%201.5%5Ctimes%2010%5E-9C%5Ctimes%201.1%5Ctimes%2010%5E-8C%7D%7B%28.43-x%29%5E2%7D)
![\frac{5\times 10^-9C}{x^2}=\frac{1.5\times 10^-9C}{(.43-x)^2}](https://tex.z-dn.net/?f=%5Cfrac%7B5%5Ctimes%2010%5E-9C%7D%7Bx%5E2%7D%3D%5Cfrac%7B1.5%5Ctimes%2010%5E-9C%7D%7B%28.43-x%29%5E2%7D)
![5\times 10^-9C \times (.43-x)^2=1.5\times 10^-9C\times (x)^2](https://tex.z-dn.net/?f=5%5Ctimes%2010%5E-9C%20%5Ctimes%20%28.43-x%29%5E2%3D1.5%5Ctimes%2010%5E-9C%5Ctimes%20%28x%29%5E2)
![3.5x^2-4.3x+.9245=0](https://tex.z-dn.net/?f=3.5x%5E2-4.3x%2B.9245%3D0)
on solving we get
x=.277m
0r x=28cm
The distance between charge
and
, is 28cm
Answer:
The same amount of energy is required to either stretch or compress the spring.
Explanation:
The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:
![U=\frac{1}{2}k (\Delta x)^2](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7Dk%20%28%5CDelta%20x%29%5E2)
where
k is the spring constant
is the stretch/compression of the spring
In the first case, the spring is stretched from x=0 to x=d, so
![\Delta x = d-0=d](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%20d-0%3Dd)
and the amount of energy required is
![U=\frac{1}{2}k d^2](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7Dk%20d%5E2)
In the second case, the spring is compressed from x=0 to x=-d, so
![\Delta x = -d -0 = -d](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%20-d%20-0%20%3D%20-d)
and the amount of energy required is
![U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7Dk%20%28-d%29%5E2%3D%20%5Cfrac%7B1%7D%7B2%7Dkd%5E2)
so we see that the amount of energy required is the same.
Answer:
Explanation:
95.0 km/hr = 26.39 m/s
65 km/hr = 18.06 m/s
Circumference of a tire is 0.9π m
77 revolutions is a distance of
77(0.9π) = 69.3π m
v² = u² + 2as
a = (v² - u²) / 2s
a = (18.06² - 26.39²) / (2(69.3π))
a = -0.85 m/s²
s = (v² - u²) / 2a
s = (0² - 26.39²) / 2(-0.85)
s = 409 m