Answer:

Explanation:
First, the instant associated to the angular displacement is:

Roots of the second-order polynomial are:

Only the first root is physically reasonable.
The angular velocity is obtained by deriving the angular displacement function:


The angular acceleration is obtained by deriving the previous function:

The resultant linear acceleration on the rim of the disk is:






Answer:
<em>171.5m</em>
Explanation:
The velocity of sound in water = 343m/s
Time taken = 1.00secs
using the formula to calculate the distance
2x = vt
x is the distance
v is the speed of sound
t is the time
x = vt/2
x = 343(1)/2
x = 171.5m
<em>hence their separation 1.00 s after the second object is released is 171.5m</em>
By conducting exhaustive, high-intensity online research for about 15 seconds, I found a source that says the speed of sound in copper is 4600 m/s. (You could easily have completed the same research project in about 1/3 of the time it took you to type and post the question here.)
Time it takes = (distance) / (speed)
Time = (25,000 meters) / (4600 m/s)
Time = (25 / 4.6) km-sec/km
<em>Time = 5.43 seconds </em>
The answer is 0.000824653J
You need to use the formula Mass * Velocity^2 over 2
Answer:
Coil 2 have 235 loops
Explanation:
Given
The number of loops in coil 1 is n
₁=
159
The emf induced in coil 1 is ε
₁
=
2.78
V
The emf induced in coil 2 is ε
₂
=
4.11
V
Let
n
₂ is the number of loops in coil 2.
Given, the emf in a single loop in two coils are same. That is,
ϕ
₁/n
₁=
ϕ
₂
n
₂⟹
2.78/159
=
4.11/
n
₂
n₂=
n₂=235
Therefore, the coil 2 has n
₂=
235 loops.