Explanation:
Given that,
Height of object = 3.0 cm
Distance of object u= 48 cm
Focal length = 48 cm
We need to calculate the image distance
Using formula of lens

Put the value into the formula




(I). The image is real.
(II). The distance of the image from the lens is 14.11 cm.
(II). The image is inverted.
(IV). We need to calculate the height of the image
Using formula of magnification


Put the value into the formula



The height of the image is 0.88 cm.
Hence, This is the required solution.
For any temperature scale the ratio of difference of temperature will be same
so we can write

now given that in both scales the temperature must be same




<em>so at above temperature both scales will have same temperature</em>
Answer:
The force of weight and drag force of the air
Explanation:
This is because when the ball leaves the hand the only force acting on it is the gravity of the earth, therefore its weight, since the force applied by the quarterback is no longer being applied to the ball. You can also consider the drag force of the air in the ball but in some cases it is minimal and can be neglected.
Answer : 312 meters/second.
Explanation : As the given expression is v=dt−v0;
where v is to be found and rest all values are given;
d is 35 meters
t is 9 seconds
and v0 is 3 seconds
So, when we substitute these values in the equation we get,
v = (35 meters X 9 seconds) - 3 seconds
v = 312
therefore, velocity v will be equal to 312.
The total flux through the cylinder is zero.
In fact, the electric flux through a surface (for a uniform electric field) is given by:

where
E is the intensity of the electric field
A is the surface
is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.
We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because
and
).
On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be

where r is the radius, and where we have taken
since the perpendicular to the surface is parallel to the direction of the electric field, so
. On the second surface, however, the perpendicular to the surface is opposite to the electric field, so
and
, therefore the flux is

And the net flux through the cylinder is
