An electron beam moves toward a cathode ray tube screen, which is 50 cm away from the negative electrode. The electrons are acce
lerated by a potential difference of 18 kV. Estimate the maximum displacement of the electron beam caused by Earth's magnetic field. The average magnetic field at the surface of Earth is roughly BE = 45×10−6T. Δz = ?
1 answer:
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Answer:
The force of static friction is 17.15 N
Explanation:
It is given that,
Mass of the bucket, m = 5 kg
The coefficient of static friction is,
We need to find the maximum force of static friction. It is given by :
F = 17.15 N
So, the force of static friction is 17.15 N. Hence, this is the required solution.
Answer:
A) v = 1,675 10³ m / s , B) r₂ = 11,673 10⁶ m
Explanation:
A) This exercise we must use Newton's second law, where the forces of gravity are the Moon
F = m a
acceleration is centripetal
a = v² / r
force is the force of universal attraction
F = G m M / r²
we substitute
G m M / r² = m v² / r
v² = G M / r
distance
r = R_moon + h
r = 1.74 10⁶ +1.0786 10⁴
r = 1,750786 10⁶ m
we calculate
v = √ (6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶)
v = √ (2,8052 10⁶)
v = 1,675 10³ m / s
B) let's use energy conservation
Starting point. In the mountain
Em₀ = K + U = ½ m v² + G m M / r
Final point. Where the speed is zero
= U = G mM / r₂
Em₀ = Em_{f}
½ m v² + G m M / r = G mM / r₂
1 / r₂ = (½ v₂ + G M / r) / GM
let's calculate
1 / r₂ = (½ (1,675 10³)² + 6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶) /(6.67 10⁻¹¹ 7.36 10²²)
1 / r₂ = (1,4028 10⁶ + 2,805 10⁶) / 49.12 10¹¹
1 / r₂ = 8.5664 10⁻⁷
r₂ = 11,673 10⁶ m