Answer:
0.0389 cm
Explanation:
The current density in a conductive wire is given by
where
I is the current
A is the cross-sectional area of the wire
In this problem, we know that:
- The fuse melts when the current density reaches a value of
- The maximum limit of the current in the wire must be
I = 0.62 A
Therefore, we can find the cross-sectional area that the wire should have:
We know that the cross-sectional area can be written as
where d is the diameter of the wire.
Re-arranging the equation, we find the diameter of the wire:
In longitudinal waves, the motion of the particles in a medium is across the direction of the wave.
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:
down below
Explanation:
Since we aren't the given the time, lets say that an object to 25 seconds to fall 50 meters. We can use the formula [ s = d/t ] to solve.
s = 50/25
s = 2
Therefore, the object was falling at a rate of 2 meters per second.
Best of Luck!
