Why is a cathode ray tube connected to a vacuum pump?
1 answer:
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Answer:
1. The bird close to the center
2. 4/25 of the original force.
Explanation:
1. Tangential velocity is v=w*d (in m/s), where w is the rotational speed, commonly denoted as the letter omega (in radians per second). d is the distance from the center of the rotating object to the position of where you would like to calculate the velocity (in meters).
As we can note, the furthest from the center we are calculating the velovity the higher it is, because the rotational velocity is not changing but the distance of the object with respect to the center is. If v=w*d, then the lower the d (distance) the lower the tangential velocity.
2. Take a look at the picture:
We have the basic equation for the gravitational force.
We have to forces: Fg1, which is the original force, and Fg2, the force when the mass and the distance changes.
If we consider that mass 2 didn't change (m2'=m2), mass 1 is four times its original (m1'=4*m1) and distance is 5 times the original (r'=5*r), then next step is just plugging it into the equation for Fg2.
Dividing the original force Fg1 by the new force Fg2 (notice you can just as well do the inverse, Fg2 divided by Fg1) gives us the relation between the forces, cancelling all the variables and being left only with a simple fraction!
Answer:
The near point of an eye with power of +2 dopters, u' = - 50 cm
Given:
Power of a contact lens, P = +2.0 diopters
Solution:
To calculate the near point, we need to find the focal length of the lens which is given by:
Power, P =
where
f = focal length
Thus
f =
f = = + 0.5 m
The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.
Now, by using lens maker formula:
where
u = object distance = 25 cm = 0.25 m = near point of a normal eye
u' = image distance
Now,
Solving the above eqn, we get:
u' = - 0.5 m = - 50 cm
Answer:
Explanation:
Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.
For faster car on the road,
v = 2v
..........(1)
For the slower car on the road,
............(2)
Equation (1) becomes,
So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.