What is the period that corresponds to a frequency of 39.5 Hz? Answer in units of s.
1 answer:
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>
You might be interested in
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) =
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Answer:
a)
The direction will be negative direction.
b)
The direction will be positive direction.
Explanation:
Given that
q1 = +7.7 µC is at x1 = +3.1 cm
q2 = -19 µC is at x2 = +8.9 cm
We know that electric filed due to a charge given as
Now by putting the va;ues
a)
The net electric field
The direction will be negative direction.
As we know that electric filed line emerge from positive charge and concentrated at negative charge.
b)
Now
distance for charge 1 will become =5.5 - 3.1 = 2.4 cm
distance for charge 2 will become =8.9-5.5 = 3.4 cm
The net electric field
The direction will be positive direction.