Answer:
1) The net electric field at any location inside a block of copper is zero if the copper block is in equilibrium.
2) In equilibrium, there is no net flow of mobile charged particles inside a conductor.
3) If the net electric field at a particular location inside a piece of metal is not zero, the metal is not in equilibrium.
Explanation:
1) and 3) A block of copper is a conductor. The charged particles on a conductor in equilibrium are at rest, so the intensity of the electric field at all interior points of the conductor is zero, otherwise, the charges would move resulting in an electric current.
2) The charged particles on a conductor in equilibrium are at rest.
Answer:
160 kg
12 m/s
Explanation:
= Mass of first car = 120 kg
= Mass of second car
= Initial Velocity of first car = 14 m/s
= Initial Velocity of second car = 0 m/s
= Final Velocity of first car = -2 m/s
= Final Velocity of second car
For perfectly elastic collision
Applying in the next equation
Mass of second car = 160 kg
Velocity of second car = 12 m/s
Answer:
Explanation:
The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .
Their potential energy at distance d
= kq₁q₂ / d
= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J
= 16.2 J
Their total kinetic energy will be equal to this potential energy.
2 x 1/2 x mv² = 16.2
= 3 x 10⁻⁶ v² = 16.2
v = 5.4 x 10⁶
v = 2.32 x 10³ m/s
When masses are different , total P.E, will be divided between them as follows
K E of 3 μ = (16.2 / 30+3) x 30
= 14.73 J
1/2 X 3 X 10⁻⁶ v₁² = 14.73
v₁ = 3.13 x 10³
K E of 30 μ = (16.2 / 30+3) x 3
= 1.47 J
1/2 x 30 x 10⁻⁶ x v₂² = 1.47
v₂ = .313 x 10³ m/s
The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.
Given:
Potential difference, V = 1.2 V
Charge on an electron, e = 1.6 × 10⁻¹⁹ C
Calculation:
We know that the work done to transport an electron from the positive to the negative terminal is given as:
W.D = (Charge on electron)×(Potential difference)
= e × V
= (1.6 × 10⁻¹⁹ C)×(1.2 V)
= 1.92 × 10⁻¹⁹ J
Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.
Learn more about work done on a charge here:
<u>brainly.com/question/13946889</u>
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If we use the equation:
N2 + 3H2 --> 2NH3
Then
1 mol of Nitrogen required 3 moles of Hydrogen
x mols : 6.34mols
X = 6.34/3
X = 2.11 moles of Nitrogen are required.
