Answer:
The light rays falling on a rough surface does follow the laws of reflection. The light rays are incident parallel on the rough surface but due to uneven surface the light rays are not reflected parallel rather they are reflected in different direction. Hence, no image is formed.
Answer:
Given the area A of a flat surface and the magnetic flux through the surface
it is possible to calculate the magnitude
.
Explanation:
The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux
is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (
). So 1 Wb=1 T.m².
For a flat surface S of area A in a uniform magnetic field B, with
being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

We are told the values of
and B, then we can calculate the magnitude

Answer:
8.60 g/cm³
Explanation:
In the lattice structure of iron, there are two atoms per unit cell. So:
where
an and A is the atomic mass of iron.
Therefore:

This implies that:

= 
Assuming that there is no phase change gives:

= 8.60 g/m³
] Ceres is composed of rock and ice and is estimated to comprise approximately one third of the mass of the entire asteroid belt. Ceres is the only object in the asteroid belt known to be rounded by its own gravity (though detailed analysis was required to exclude 4 Vesta). From Earth, the apparent magnitude of Ceres ranges from 6.7 to 9.3, peaking once every 15 to 16 months,[21]hence even at its brightest it is too dim to be seen with the naked eye except under extremely dark skies.
Answer:
Approximately
.
Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.
Explanation:
Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant
near the surface of the earth.
For an object that is accelerating constantly,
,
where
is the initial velocity of the object,
is the final velocity of the object.
is its acceleration, and
is its displacement.
In this case,
is the same as the change in the ball's height:
. By assumption, this ball was dropped with no initial velocity. As a result,
. Since the ball is accelerating due to gravity,
.
.
In this case,
would be the velocity of the ball just before it hits the ground. Solve for
.
.