Answer:
Maximum speed of the rod, v = 10.34 m/s
Explanation:
It is given that,
Voltage of the battery, V = 2.7 V
The magnetic field perpendicularly into the plane of the paper is, B = 0.9 T
Length of the rod between the rails, l = 0.29 m
Due to the motion of the rails inside the magnetic field, an emf will induced in it which is given by :
v is the speed attained by the rod
v = 10.34 m/s
So, the maximum speed attained by the rod after the switch is closed is 10.34 m/s. Hence, this is the required solution.
The effiency of a machine is
(output work or energy) / (input work or energy) .
For the system described in the question, that's
(123 J) / (150 J) = 0.82 = 82% .
Probably C...:):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):)
Answer:
a) 7200 ft/s²
b) 140 ft
c) 3.7 s
Explanation:
(a) Average acceleration is the change in velocity over change in time.
a_avg = Δv / Δt
We need to find what velocity the puck reached after it was hit by the hockey player.
We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s². Therefore:
v² = v₀² + 2a(x − x₀)
(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)
v₀² = 5200 ft²/s²
v₀ = 20√13 ft/s
So the average acceleration impacted to the puck as it is struck is:
a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)
a_avg = 2000√13 ft/s²
a_avg ≈ 7200 ft/s²
(b) The distance the puck travels before stopping is:
v² = v₀² + 2a(x − x₀)
(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)
x = 140 ft
(c) The time the puck takes to travel 10 ft without friction is:
t = (10 ft) / (20√13 ft/s)
t = (√13)/26 s
The time the puck travels over the rough ice is:
v = at + v₀
(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)
t = √13 s
So the total time is:
t = (√13)/26 s + √13 s
t = (27√13)/26 s
t ≈ 3.7 s
Answer:
Force = 8.0 k g m / s
Explanation:
Force = mass x acceleration
Mass = 4.0 k g Acceleration = 2.0 m / s 2
Hence,force = ( 4.0 x 2.0 ) k g m / s 2 = 8.0 k g m / s 2
