Answer:

Explanation:
Given data:
Rotating cylinder length = 9 mi
diameter of cylinder is 5.9 mi
we know that linear acceleration is given as
a = r ω^2
where ω is angular velocity
so



Answer: 1m/s
Explanation: according to the law of conservation of linear momentum in an isolated system, the momentum of the gun equals that of the bullet.
Mathematically
Mb×Vb = Mg×Vg
Where Mb = mass of bullet = 1/100 = 0.01 kg
Vb = velocity of bullet = 200 m/s
Mg = mass of gun = 2kg
Vg = recoil velocity of gun =?
0.01×200 = 2×Vg
Vg = 0.01×200/2
Vg = 0.01×100
Vg = 1m/s
Answer:
A. The resonator behaves as a wave guide (a hollow pipe used as a transmission line). The characteristics of the pipe depend on the type of the wave to be transmitted.
We have to add two vectors.
Vector #1: 0.15 m/s north
Vector #2: 1.50 m/s east
Their sum:
Magnitude: √(0.15² + 1.50²)
Magnitude = √(0.0225+2.25)
Magnitude = √2.2725
Magnitude = <em>1.5075 m/s</em>
Direction = arctan(0.15/1.50) north of east
Direction = <em>5.71° north of east</em>
Answer:
Wave A
<em>I</em><em> </em><em>hope this</em><em> </em><em>helps</em><em> </em>