Answer with Explanation:
We are given that
Diameter of fighter plane=2.3 m
Radius=
a.We have to find the angular velocity in radians per second if it spins=1200 rev/min
Frequency=
1 minute=60 seconds
Angular velocity=
Angular velocity=
b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

c.Centripetal acceleration=
Centripetal acceleration==
Answer:
It's a strength or force
Explanation:
the quantity of motion of a moving body, measured as a product of its mass and velocity
Answer:
v = 1.98*10^8 m/s
Explanation:
Given:
- Rod at rest in S' frame
- makes an angle Q = sin^-1 (3/5) in reference frame S'
- makes an angle of 45 degree in frame S
Find:
What must be the value of v if as measured in S the rod is at a 45 degree)
Solution:
- In reference frame S'
x' component = L*cos(Q)
y' component = L*sin(Q)
- Apply length contraction to convert projected S' frame lengths to S frame:
x component = L*cos(Q) / γ (Length contraction)
y component = L*sin(Q) (No motion)
- If the rod is at angle 45° to the x axis, as measured in F, then the x and y components must be equal:
L*sin(Q) = L*cos(Q) / γ
Given: γ = c / sqrt(c^2 - v^2)
c / sqrt(c^2 - v^2) = cot(Q)
1 - (v/c)^2 = tan(Q)
v = c*sqrt( 1 - tan^2 (Q))
For the case when Q = sin^-1 (3/5)::
tan(Q) = 3/4
v = c*sqrt( 1 - (3/4)^2)
v = c*sqrt(7) / 4 = 1.98*10^8 m/s
Answer:
The velocity of the block is 231650.8 ft/s
Explanation:
Knowing
W = 10 lb
v0 = u = 4 ft/s
F = 8
lb
ms = 0.2
Applying the newton second law
ΣFy = 0
N - W = 0 --> N = W = 10 lb
ΣFx = m
F - ms N = ma
- 0.2(10) = 
a = 
Using Kinematics
a = dv/dt
=
= 
v - 4 = 3.22 
when t = 30
v = 231650.8 ft/s
Answer
given,
frequency = 523 Hz
piano tuner hears = 2.00 beats/s
a) possible frequency = 523 ± 2
525 Hz and 521 Hz
b) On tightening piano wire, tension will be increased which will cause an increase of frequency of piano wire. Since no. of beats heard is also increased, the frequency of piano wire must be greater than the frequency of oscillator.
f p' = 5+ f₀
f p' = 3 + 523 Hz = 526 Hz.
c) f = k x √T
Let the tension at 523 Hz be T₀ and at 526 Hz it be T₁
523 = k x √T₀
T₀ = 523²/ k²
T₁ = 526²/ k²
% change in tension
= 
=
= 1.13 %