The answer for the following answer is answered below.
- <u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>
- <u><em>Therefore the option for the answer is "B".</em></u>
Explanation:
Frequency (f):
The number of waves that pass a fixed place in a given amount of time.
The SI unit of frequency is Hertz (Hz)
Time period (T):
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) = 100 Hz
wavelength (λ) = 2.0 m
To calculate:
Time period (T)
We know;
According to the formula;
<u>f =</u>
<u></u>
Where,
f represents the frequency
T represents the time period
from the formula;
T = 
T = 
T = 0.01 seconds
<u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>
I think is True! Is the best answer. Because the trumpet it make them sounds like the lips with the musician and it vibrate.
where:
F - force
m - mass
a - acceleration
We transform this formula to get a:
Answer:
Approximately 
. (Assuming that the drag on this ball is negligible, and that 
.)
Explanation:
Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:
- Horizontal: no acceleration, velocity is constant (at
is constant throughout the descent.) - Vertical: constant downward acceleration at , starting at
.
The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: 
. Combine these two quantities to find the duration of this descent:
.
In other words, the ball in this question start at a vertical velocity of 
, accelerated downwards at , and reached the ground after 
.
Apply the SUVAT equation 
to find the vertical displacement of this ball.
![\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%20x%28%5Ctext%7Bvertical%7D%29%20%5C%5C%5B0.5em%5D%20%26%3D%20-%5Cfrac%7B1%7D%7B2%7D%5C%2C%20g%20%5Ccdot%20t%5E%7B2%7D%20%2B%20v_0%5Ccdot%20t%5C%5C%5B0.5em%5D%20%26%3D%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2010%5C%3B%20%5Crm%20m%20%5Ccdot%20s%5E%7B-2%7D%20%5Ctimes%20%287.5%5C%3B%20%5Crm%20s%29%5E%7B2%7D%20%5C%5C%20%26%20%5Cquad%20%5Cquad%20%2B%200%5C%3B%20%5Crm%20m%20%5Ccdot%20s%5E%7B-1%7D%20%5Ctimes%207.5%5C%3B%20s%20%5C%5C%5B0.5em%5D%20%26%3D%20-281.25%5C%3B%20%5Crm%20m%5Cend%7Baligned%7D)
.
In other words, the ball is below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be 
.
PART a)
here when stone is dropped there is only gravitational force on it
so its acceleration is only due to gravity
so we will have
Part b)
Now from kinematics equation we will have
now we have
y = 25 m
so from above equation
Part c)
If we throw the rock horizontally by speed 20 m/s
then in this case there is no change in the vertical velocity
so it will take same time to reach the water surface as it took initially
So t = 2.26 s
Part D)
Initial speed = 20 m/s
angle of projection = 65 degree
now we have
PART E)
when stone will reach to maximum height then we know that its final speed in y direction becomes zero
so here we can use kinematics in Y direction
so it will take 1.85 s to reach the top
