Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

v is the speed of cat, 

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
I want to say that they will be primarily flat but I honestly don't know
PART A)
By Snell's law we know that

here we know that



now from above equation we have


so it will refract by angle 39.3 degree
PART B)
Here as we can see that image formed on the other side of lens
So it is a real and inverted image
Also we can see that size of image is lesser than the size of object here
Here we can use concave mirror to form same type of real and inverted image
PART C)
As per the mirror formula we know that



so image will form at 30 cm from mirror
it is virtual image and smaller in size
Answer:3,600 Newtons
Explanation:
The net force acting on the car is
3×10^3squared
Newtons.
Force is defined as the product of the mass of the body and its aaceleration,⇒F=ma
Substituting the above given values we get,F=(1500kg) (2.0m /s^2 squared)=3000 N=3×10^3 squared N.
N=newtons
Answer:
the ans will be because it has 1.672