Answer:
a = αR
Then We apply the Newton's second law of motion:
∑ F = ma
F - mg sinθ - f = ma.
F - mg sinθ - μmg cosθ = ma
Using given data we have:
F - 3217 = 470a ........................... (1)
Apply the Newton's law for rotation:
∑τ = I α
FR - (mg sinθ)R = 
then: F - mg sinθ = 3ma / 2.
F - 2774 = 705a .......................... (2)
solving 1 and 2 we get:
F = 4103 N
a = 1.885 m/s²
b) In this case the time is given by:
t = 
it comes from: d = 
starting from rest vo = 0
t = 
= 3.09 s
Answer:
The frequency of the waves depends on the distance between wave fronts - considering a front as a maximum disturbance of the wave
(Consider the waves emitted by an organ pipe: condensation and rarefactions)
The waves themselves are a fixed distance apart -
as one moves towards the source the waves received will be closer together (higher frequency)
So if the frequency received increases, the distance between the source and the observer must be decreasing
Answer:
a) 096V b) 0.0288A c) 0.3456W
Explanation:
a) Vp/Vs= Np/Ns
120/Vs= 500/4
Vs= 096V
b) Np/Ns= Is/Ip
500/4= 3.6/Ip
Ip= 0.0288A
c) P= VI
P=(120)(0.0288)
P= 0.3456W
34 + 3.7395 = 37.7395
The answer is 38.
The answer is written to two significant figures because the smallest given number of significant figure is 2.
Answer:
The tension in string will be "3.62 N".
Explanation:
The given values are:
Length of string:
l = 3 ft
or,
= 0.9144 m
frequency,
f = 60 Hz
Weight,
= 0.096 lb
or,
= 0.0435 kgm/s²
Now,
The mass will be:
= 
= 
As we know,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
or,
⇒ 
⇒ 
⇒ 
Now,
⇒ 
or,
⇒ 
On putting the above given values, we get
⇒ 
⇒ 
⇒ 
