The answer for the following problem is mentioned below.
- <u><em>Therefore the time period is 0.02 seconds.</em></u>
Explanation:
Frequency:
The number of waves that pass a fixed place in a given amount of time. (or)
The number of waves that pas by per second.
The SI unit of the frequency is Hertz(Hz).
Time period:
The time taken for one complete cycle of vibration to pass a given point.
The SI unit of time period is seconds. (s)
Given:
Frequency (f) = 39.5 Hz
To calculate:
Time period (T)
We know;
According to the problem;
From the problem;
<u>f = </u>
<u></u>
Where;
f represents the frequency
T represents the time period
f = 
f = 0.02 seconds
<u><em>Therefore the time period is 0.02 seconds.</em></u>
Answer:
Saturn's differential rotation will cause the length of a day measures to be longer by 0.4 hours
Explanation:
Differential rotation occurs due to the difference in angular velocities of an object as we move along the latitude of the or as we move into different depth of the object, indicating the observed object is in a fluid form
Saturn made almost completely of gas and has differential motion given as follows
Rotation at the equator = 10 hours 14 minutes
Rotation at high altitude = 10 hours 38 minutes
Therefore;
The differential rotation = 10 hours 38 minutes - 10 hours 14 minutes
The differential rotation = 24 minutes = 24 minutes × 1 hour/(60 minutes) = 0.4 hours
The differential rotation = 0.4 hours
Therefore, the measured day at the higher altitude will be 0.4 longer than at the equator.
Answer:
Answer:u=66.67 m/s
Explanation:
Given
mass of meteor m=2.5 gm\approx 2.5\times 10^{-3} kg
velocity of meteor v=40km/s \approx 40000 m/s
Kinetic Energy of Meteor
K.E.=\frac{mv^2}{2}
K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}
K.E.=2\times 10^6 J
Kinetic Energy of Car
=\frac{1}{2}\times Mu^2
=\frac{1}{2}\times 900\times u^2
\frac{1}{2}\times 900\times u^2=2\times 10^6
900\times u^2=4\times 10^6
u^2=\frac{4}{9}\times 10^4
u=\frac{2}{3}\times 10^2
u=66.67 m/s
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = 
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
