Answer:
Explanation:
mass of string = .0125 / 9.8
= 1.275 x 10⁻³ kg
Length of string l = 1.5 m .
m = mass per unit length
= ( .1.275 / 1.5) x 10⁻³ kg/m
m = .85 x 10⁻³ kg/m
wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)
compare with equation of wave
y(x,t) = Acos(K x − ω t)
ω ( angular velocity ) = 4830 rad/s
k = 172 rad/m
Velocity = ω / k
= 4830/172 m /s
= 28.08 m /s
velocity of wave = 
28.08 = 
788.48 = W / .85 X 10⁻³
W = 670 x 10⁻³ N .
c ) wave length
wave length =2π / k
= 2 x 3.14 / 172
= .0365 m
no of wave lengths over whole length of string
= 1.5 / .0365
= 41
d )
equation for waves traveling down the string
= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)
HEJEGEJDHIEBDJDHDIDGDJGJDHD
V=IR
60-V
The current that passes through a 10-ohm resistor = I
I=60/10
6 amperes
The answer is
C. periodic fluctuations in the intensity if sound waves.
Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
a = -1.47 m/ (a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/ (since same friction force is applied)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
