Answer:
horizontal component of normal force is equal to the centripetal force on the car
Explanation:
As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle
This force is due to friction force when car is moving in circle with uniform speed
Now it is given that car is moving on the ice surface such that the friction force is zero now
so here we can say that centripetal force is due to component of the normal force which is due to banked road
Now we have
![N sin\theta = \frac{mv^2}{R}](https://tex.z-dn.net/?f=N%20sin%5Ctheta%20%3D%20%5Cfrac%7Bmv%5E2%7D%7BR%7D)
![N cos\theta = mg](https://tex.z-dn.net/?f=N%20cos%5Ctheta%20%3D%20mg)
so we have
![v = \sqrt{Rg tan\theta}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7BRg%20tan%5Ctheta%7D)
so this is horizontal component of normal force is equal to the centripetal force on the car
Answer:
n=2.053
Explanation:
We will use Snell's Law defined as:
![n_{1}*Sin\theta_{1}=n_{2}*Sin\theta_{2}](https://tex.z-dn.net/?f=n_%7B1%7D%2ASin%5Ctheta_%7B1%7D%3Dn_%7B2%7D%2ASin%5Ctheta_%7B2%7D)
Where n values are indexes of refraction and
values are the angles in each medium. For vacuum, the index of refraction in n=1. With this we have enough information to state:
![1*Sin(39.9)=n_{2}*Sin(18.2)](https://tex.z-dn.net/?f=1%2ASin%2839.9%29%3Dn_%7B2%7D%2ASin%2818.2%29)
Solving for
yields:
![n_{2}=\frac{Sin(39.9)}{Sin(18.2)}=2.053](https://tex.z-dn.net/?f=n_%7B2%7D%3D%5Cfrac%7BSin%2839.9%29%7D%7BSin%2818.2%29%7D%3D2.053)
Remember to use degrees for trigonometric functions instead of radians!
Answer:
7.04 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity = 0
s = Displacement on Earth = 1.2 m
a = Acceleration due to gravity on Moon = 1.67 m/s²
a = Acceleration due to gravity Earth= 9.81 m/s²
Accelration going up is considered as negetive
Initial Velocity of the ball
![v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as%5C%5C%5CRightarrow%20-u%5E2%3D2as-v%5E2%5C%5C%5CRightarrow%20-u%5E2%3D2%5Ctimes%20-9.81%5Ctimes%201.2-0%5E2%5C%5C%5CRightarrow%20u%3D%5Csqrt%7B2%5Ctimes%209.81%5Ctimes%201.2%7D%5C%5C%5CRightarrow%20u%3D4.85%5C%20m%2Fs)
Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is
![v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as%5C%5C%5CRightarrow%20s%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2a%7D%5C%5C%5CRightarrow%20s%3D%5Cfrac%7B0%5E2-4.85%5E2%7D%7B2%5Ctimes%20-1.67%7D%5C%5C%5CRightarrow%20s%3D7.04%5C%20m)
The displacement of the ball on the moon is 7.04 m
As AL2006 correctly pointed out the formula is 1/2 kx^2. I was thinking of force and work is the integral of force over the distance applied. So now
![W= \frac{1}{2}k x^{2}](https://tex.z-dn.net/?f=W%3D%20%5Cfrac%7B1%7D%7B2%7Dk%20x%5E%7B2%7D%20%20)
and
Metals tend to readily lose electrons and form cations. Most of them react with atmospheric oxygen to form metal oxides. However, different metals have different reactivities towards oxygen (unreactive metals such as gold and platinum do not readily form oxides when exposed to air