A 65 kg bungee jumper jumps off of a bridge. At an instant when they are 50 m above the ground, they are falling at a speed of 4
2 answers:
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The mass of the quarterback is 61.2 kg.
Explanation:
mass of the football player = m1 = 102 kg
mass of the quarterback = m2 = ?
velocity of the football player = v1 = 8 m/s
According to the law of conservation of momentum:
The total momentum of a system before and after the collision remains constant. Assuming the situation as an isolated system which is not affected by any external factors, we have:
m₁v₁ + m₂v₂ = (m₁+m₂)V
Here, we need to find m₂.
We assume that the quarterback is standing still when he is attacked by the football player so v₂ = 0 m/s
After the collision both of them fall to the ground with a velocity of 5 m/s so V = 5 m/s
Keywords: momentum, velocity, law of conservation of momentum
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Answer:
1) The total charge of the top plate is 0.008 C
b) The total charge of the bottom plate is -0.008 C
2) The electric field at the point exactly midway between the plates is 0
3) The electric field between plates is approximately 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N
Explanation:
The given parameters of the parallel plate capacitor are;
The dimensions of the plates = 4 × 2 cm
The distance between the plates = 10 cm
The surface charge density of the top plate, σ₁ = 10 C/m²
The surface charge density of the bottom plate, σ₂ = -10 C/m²
The surface area, A = 0.04 m × 0.02 m = 0.0008 m²
1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C
b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C
2) The electrical field at the point exactly midway between the plates is given as follows;
Therefore, we have;
The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m
The electric field at the point exactly midway between the plates, = 0
3) The electric field, 'E', between plates is given as follows;
E ≈ 1.1294 × 10¹² N/C
The electric field between plates, E ≈ 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates
The charge on an electron, e = -1.6 × 10⁻¹⁹ C
The force on an electron in the middle of the two plates, = E × e
∴ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N
The force on an electron in the middle of the two plates, ≈ 1.807 × 10⁻⁷ N