Answer:
Images formed by a convex mirror are always virtual
Explanation:
A virtual image is always created by a convex mirror, and it is always situated behind the mirror. The picture is vertical and situated at the focus point when the item is far away from the mirror. As the thing approaches the mirror, the image follows suit and increases until it reaches the same height as the object.
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Answer:
, assuming that the gravitational field strength is
.
Explanation:
Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.
By Newton's Second Law, the net force on this block would be
. External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.
Let
denote the mass of this block. It is given that
. The weight of this block would be:
.
Hence, the force that the boy applies on this block would be upward with a magnitude of
.
The mechanical work that a force did is equal to the product of:
- the magnitude of the force, and
- the displacement of the object in the direction of the force.
The displacement of this block (upward by
) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of:
- the magnitude of the force that this boy exerted,
, and - the displacement of this block in the direction,
.
.
Answer
given,
heat added to the gas,Q = 3300 kcal
initial volume, V₁ = 13.7 m³
final volume, V₂ = 19.7 m³
atmospheric pressure, P = 1.013 x 10⁵ Pa
a) Work done by the gas
W = P Δ V
W = 1.013 x 10⁵ x (19.7 - 13.7)
W = 6.029 x 10⁵ J
b) internal energy of the gas = ?
now,
change in internal energy
Δ U = Q - W
Q = 3300 x 10³ cal
Q = 3300 x 10³ x 4.186 J
Q = 1.38 x 10⁷ J
now,
Δ U = 1.38 x 10⁷ - 6.029 x 10⁵
Δ U = 1.32 x 10⁷ J
To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.
The drift velocity is given by the equation:

Where
I = current
n = Number of free electrons
A = Cross-Section Area
q = charge of proton
Our values are given by,






The hall voltage is given by

Where
B= Magnetic field
n = number of free electrons
d = distance
e = charge of electron
Then using the formula and replacing,


Answer:
it's D. Make the column wider
Explanation: