Answer:
Average speed=1.5 m/s
Frequency of pendulum=93.75Hz
Explanation:
We are given
Frequency, 
Average wavelength =
Speed of pendulum, 
Wavelength, 
We have to find the average speed and frequency of pendulum.
We know that
Speed,
Using the formula
Average speed,
Hence, the average speed =1.5m/s
Frequency, 
Using the formula


Hence, the frequency of a pendulum=93.75Hz
Answer:
frequency is 195.467 Hz
Explanation:
given data
length L = 4.36 m
mass m = 222 g = 0.222 kg
tension T = 60 N
amplitude A = 6.43 mm = 6.43 ×
m
power P = 54 W
to find out
frequency f
solution
first we find here density of string that is
density ( μ )= m/L ................1
μ = 0.222 / 4.36
density μ is 0.050 kg/m
and speed of travelling wave
speed v = √(T/μ) ...............2
speed v = √(60/0.050)
speed v = 34.64 m/s
and we find wavelength by power that is
power = μ×A²×ω²×v / 2 ....................3
here ω is wavelength put value
54 = ( 0.050 ×(6.43 ×
)²×ω²× 34.64 ) / 2
0.050 ×(6.43 ×
)²×ω²× 34.64 = 108
ω² = 108 / 7.160 ×
ω = 1228.16 rad/s
so frequency will be
frequency = ω / 2π
frequency = 1228.16 / 2π
frequency is 195.467 Hz
Answer: Final speed
Explaination: because its final.
The speed of an object can be determined from the distance vs time graph.
You know that speed = distance/time
in the graph, distance/time = slope of the curve.
So SPEED IS GIVEN BY THE SLOPE of the curve in the graph.
● If the distance vs time curve is a straight line, parallel to time axis(x-axis), slope is 0. That means speed is 0. So the object is at rest.
● If the distance vs time curve is a straight line, with some non-zero slope; That means speed is nonzero and constant. So the object is in uniform motion.
● If the distance vs time curve is a curved, the slope is changing. That means speed is changing. So the object is in an accelerated motion.
Answer:
121.3 cm^3
Explanation:
P1 = Po + 70 m water pressure (at a depth)
P2 = Po (at the surface)
T1 = 4°C = 273 + 4 = 277 K
V1 = 14 cm^3
T2 = 23 °C = 273 + 23 = 300 K
Let the volume of bubble at the surface of the lake is V2.
Density of water, d = 1000 kg/m^3
Po = atmospheric pressure = 10^5 N/m^2
P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2
Use the ideal gas equation

By substituting the values, we get

V2 = 121.3 cm^3
Thus, the volume of bubble at the surface of lake is 121.3 cm^3.