By definition;
M = fo/fe
Where,
M = Angular magnification
fo = Focal length of objective lens
fe = Focal length of eyepiece lens
From the information given;
M = 180/30 = 6
Answer:
A 50 kg ball traveling at 20 m/s would have 4 times more kinetic energy.
A 50 kg ball traveling at 5 m/s would have 4 times less kinetic energy.
A 50 kg person falling at 10 m/s would have the same kinetic energy.
Explanation:
hope this helps:)
Initially its moving with tail wind so here the speed of wind will support the motion of the plane
so we can say



now when its moving with head wind we can say that wind is opposite to the motion of the plane



now by using above two equations we can find speed of palne as well as speed of wind


Answer:

Explanation:
The electric flux through a certain surface is given by (for a uniform field):

where:
E is the magnitude of the electric field
A is the area of the surface
is the angle between the direction of the field and of the normal to the surface
In this problem, we have:
is the electric field
L = 2.0 m is the side of the sheet, so the area is

, since the electric field is perpendicular to the surface
Therefore, the electric flux is

Answer:
Disruption to electricity power grid
Explanation:
We're looking a a solar flare. This will whip solar particles at high velocity into space and, If they are near earth, will interact with the earth's magnetic field. These magnetic changes will be measurable in the electric grid. Whether they are strong enough to cause "disruption" depends on a huge number of factors such as strength of and angles of the interacting magnetic fields and location of grid infrastructure,