Answer:
Two off-centered spots in the first phase of the experiment; one centered spot in the second phase of the experiment.
Explanation:
If two particles are selected in which both have the same electron mass and the same velocity, but one of the particles has a charge and the other particle has a charge of 2e. During the first stage of the experiment, the two particles have an electric force equal to F = Eq in the entire vertical direction. The acceleration of particle is equal to a = (Eq)/m.
In the second part of the experiment, the magnetic field cancels the electric field. In this way, the electric force and the magnetic force cancel each other out. Therefore, the net force acting on each particle is equal to zero.
Because these two forces cancel each other out, the particles fail to create two off-center points on the screen in the second part of the experiment. Also, if the loads are different, the deviation is also different. In this way, an off-center point cannot be achieved in the first part of the experiment. There will be two off-center points.
Answer:
The lifeguard takes 25.9 seconds to reach the child, at 25.9 meters from the start point downstream.
Explanation:
As the image shows, the child trajectory, the lifeguard trajectory and the distance from the bank form a triangle. This triangle is formed by the distances, an we already know the distance from the bank and the speed of child, and the speed of the lifeguard. So we have unknom time in common. Lets see the equations:
Using phitagoras theorem
Answer:
it looks like dots and just black space on a large scale
Explanation:
on a large scale the universe especially our milky way looks small
hope this helps
Answer: 45.95 m/s
Explanation:
When we talk about circular motion, the object's <u>acceleration</u> 
(which is a vector quantity) has two components: the <u>centripetal acceleration</u> 
always directed to the center of the circular track and the <u>tangential acceleration</u> 
which is tangent to the circular path.
Since both vectors are perpendicular to each other, the magnitude of can be calculated by the Pithagorean Theorem:
(1)
Where:
where 
is the speed and 
is the radius of the circle
can be calculated knowing the initial speed (
) and final speed (
) of the car and the time (
) it takes to accelerate at this constant rate:
Rewritting (1):
(2)
Isolating :
(3)
(4)
Finally:
