Answer:
Explanation:
<u>Given Data:</u>
Length = l = 820 mm = 0.82 m
Acceleration due to gravity = g = 9.8 ms⁻²
<u>Required:</u>
Frequency = f = ?
<u>Formula:</u>
<u>Solution:</u>
![\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%20%3D%5Cfrac%7B1%7D%7B2%20%5Cpi%7D%20%5Csqrt%7B%5Cfrac%7Bg%7D%7Bl%7D%20%7D%20%5C%5C%5C%5CPut%5C%20the%5C%20givens%5C%5C%5C%5Cf%3D%5Cfrac%7B1%7D%7B2%20%5Cpi%7D%20%5Csqrt%7B%5Cfrac%7B9.8%7D%7B0.82%7D%20%7D%5C%5C%5C%5C%20f%20%3D%200.159%20%5Ctimes%20%5Csqrt%7B11.95%7D%20%5C%5C%5C%5Cf%3D0.159%20%5Ctimes%203.457%5C%5C%5C%5Cf%3D0.55%20%5C%20Hz%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Answer:
1.549 m
Explanation:
Given:
The radius of the circular board, r = 2 m
The probability of hitting the red is given as 0.6
Now, this probability of hitting the red can be conclude as
0.6 = (Area of red)/ (Total area of the board)
Total area of the board = πr² = π × 2²
let the radius of the red area be R
thus, area of red circle, = πR²
on substituting the value of the area, we have
0.6 = (πR²)/ (π × 2²)
or
R² = 2.4
or
R = 1.549 m
Thus, the radius of the red circle is 1.549 m
The study of sound is called sonics and the study of sound waves are acoustics
Answer:
The ball would hit the floor approximately 
after leaving the table.
The ball would travel approximately 
horizontally after leaving the table.
(Assumption: 
.)
Explanation:
Let 
denote the change to the height of the ball. Let 
denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let 
denote the initial vertical velocity of this ball.
If the air resistance on this ball is indeed negligible:
.
The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was 
.
The height of the table was 
. Therefore, after hitting the floor, the ball would be 
below where it was before leaving the table. Hence, 
.
The equation becomes:
.
Solve for :
.
In other words, it would take approximately for the ball to hit the floor after leaving the table.
Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at 
) until the ball hits the floor.
The ball was in the air for approximately 
and would have travelled approximately 
horizontally during the flight.
