<span>When the two pucks collide, momentum will still be same but it acts in two different directions.
Momentum formula goes like this M = m x v, m is mass and v is speed. That means velocity is inversely proportional to mass.
So the puck with less weight changes the most, so the 0.156 kg puck changes velocity the most
.</span>
1 growth. Go from one cell/( zygote to a trillion)
2 replace. Repair\ 50 million cells die second.
3 reproduction. ( make cells for reproduction make specialized sex cells)
Answer:
Yes. Towards the center. 8210 N.
Explanation:
Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.
In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.
The net force is equal to ![F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bmv%5E2%7D%7BR%7D%20%3D%20%5Cfrac%7B1100%5Ctimes%20%2826.3%29%5E2%7D%7B68%7D%20%3D%201.1%5Ctimes%2010%5E4~N)
Note that 95 km/h is equal to 26.3 m/s.
This is the centripetal force and equal to the x-component of the applied force.
![F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3](https://tex.z-dn.net/?f=F%20%3D%20mg%5Csin%2816%29%20%3D%201100%289.8%29%5Csin%2816%29%20%3D%202.97%5Ctimes10%5E3)
As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.
The amount of the friction force should be ![8.21\times 10^3~N](https://tex.z-dn.net/?f=8.21%5Ctimes%2010%5E3~N)
Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.
Answer:
The surface charge density on planes A and B respectively is
![\sigma__{A}} } = 2\sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%202%5Csigma)
and
![\sigma__{B}} = \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BB%7D%7D%20%3D%20%20%20%5Csigma)
Explanation:
From the question we are told that
The electric field in region to the left of A is ![E_i = \frac{3 \sigma}{2 \epsilon_o}](https://tex.z-dn.net/?f=E_i%20%3D%20%20%5Cfrac%7B3%20%5Csigma%7D%7B2%20%5Cepsilon_o%7D)
The direction of the electric field is left
The electric field in the region to the right of B is ![E_f = \frac{3 \sigma}{2 \epsilon_o}](https://tex.z-dn.net/?f=E_f%20%3D%20%20%5Cfrac%7B3%20%5Csigma%7D%7B2%20%5Cepsilon_o%7D)
The direction of the electric field is right
The electric field in the region between the two planes is ![E_m = \frac{\sigma }{2 \epsilon_o }](https://tex.z-dn.net/?f=E_m%20%20%3D%20%20%5Cfrac%7B%5Csigma%20%7D%7B2%20%5Cepsilon_o%20%7D)
The direction of the electric field is right
Let the surface charge density on planes A and B be represented as ![\sigma__{A}} \ \ and \ \ \sigma__{B}} \ \ \ respectively](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%20%5C%20%5C%20and%20%20%5C%20%5C%20%5Csigma__%7BB%7D%7D%20%5C%20%5C%20%5C%20%20respectively)
From the question we see that
![E_i = E_f](https://tex.z-dn.net/?f=E_i%20%3D%20E_f)
Generally the electric to the right and to the left is due to the combined electric field generated by plane A and B so
![E_i = E_f = \frac{3\sigma }{2\epsilon} = \frac{\sigma_A }{ 2 \epsilon_o } + \frac{\sigma_B }{ 2 \epsilon_o }](https://tex.z-dn.net/?f=E_i%20%3D%20E_f%20%3D%20%20%5Cfrac%7B3%5Csigma%20%7D%7B2%5Cepsilon%7D%20%20%3D%20%20%5Cfrac%7B%5Csigma_A%20%7D%7B%202%20%5Cepsilon_o%20%7D%20%2B%20%20%5Cfrac%7B%5Csigma_B%20%7D%7B%202%20%5Cepsilon_o%20%7D)
=> ![\sigma__{A}} + \sigma__{B}} = 3 \sigma -- -(1)](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%2B%20%5Csigma__%7BB%7D%7D%20%3D%20%203%20%5Csigma%20%20--%20-%281%29)
Generally the electric field at the middle of the plane A and B is due to the diffencence in electric field generated by plane A and B
i.e
![\frac{\sigma }{2 \epsilon_o } = \frac{\sigma_A }{ 2 \epsilon_o } - \frac{\sigma_B }{ 2 \epsilon_o }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma%20%7D%7B2%20%5Cepsilon_o%20%7D%20%20%3D%20%20%5Cfrac%7B%5Csigma_A%20%7D%7B%202%20%5Cepsilon_o%20%7D%20-%20%20%20%5Cfrac%7B%5Csigma_B%20%7D%7B%202%20%5Cepsilon_o%20%7D)
=> ![\sigma__{A}} - \sigma__{B}} = \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20-%20%20%5Csigma__%7BB%7D%7D%20%3D%20%20%5Csigma)
=> ![\sigma__{A}} } = \sigma + \sigma__{B}](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%20%5Csigma%20%2B%20%5Csigma__%7BB%7D)
From equation 1
![\sigma + \sigma__{B}}+ \sigma__{B}} = 3 \sigma](https://tex.z-dn.net/?f=%5Csigma%20%2B%20%5Csigma__%7BB%7D%7D%2B%20%5Csigma__%7BB%7D%7D%20%3D%20%203%20%5Csigma)
=> ![\sigma__{B}} = \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BB%7D%7D%20%3D%20%20%20%5Csigma)
So
![\sigma__{A}} } = \sigma + \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%20%5Csigma%20%2B%20%5Csigma)
=> ![\sigma__{A}} } = 2\sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%202%5Csigma)
Answer:
1.47x10^11
Explanation:
young modulus = stress / strain
stress = force / area = 200x9.8 / 0.00004 = 49000000
strain = extension / length = 0.001 / 3 = 0.0003333
49 million / 0.0003333 = 1.47x10^11