Answer:
(a) 1.16 s
(b)0.861 Hz
Explanation:
(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.
From the question,
If 1550 cycles is completed in (30×60) seconds,
1 cycle is completed in x seconds
x = 30×60/1550
x = 1.16 s
Hence the period is 1.16 seconds.
(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).
Mathematically, Frequency is given as
F = 1/T ........................... Equation 1
Where F = frequency, T = period.
Given: T = 1.16 s.
Substitute into equation 1
F = 1/1.16
F = 0.862 Hz
Hence thee frequency = 0.862 Hz
<span>The speed of longitudinal waves, S, in a thin rod = âšYoung modulus / density , where Y is in N/m^2.
So, S = âšYoung modulus/ density. Squaring both sides, we have, S^2 = Young Modulus/ density.
So, Young Modulus = S^2 * density; where S is the speed of the longitudinal wave.
Then Substiting into the eqn we have (5.1 *10^3)^2 * 2.7 * 10^3 = 26.01 * 10^6 * 2.7 *10^6 = 26.01 * 2.7 * 10^ (6+3) = 70.227 * 10 ^9</span>
Put the object or material on a scale to figure out<span> its mass. 3. Divide the mass by the volume to </span>figure out the density<span> (p = m / v). You may also need to know </span>how to calculate<span> the volume of a </span>solid s<span>o use the formula</span>
Heat required to change the phase of ice is given by
Q = m* L
here
m = mass of ice
L = latent heat of fusion
now we have
m = 45 kg
L = 334 KJ/kg
now by using above formula
In KJ we can convert this as
so the correct answer is D option
Answer:
the force exerted on the foot by the tibia would be 2975 N
Explanation:
Given the data in the question;
To maintain equilibrium between the foot and the ball vertically, the addition normal normal force 
(750 N) and the tension in the Achilles tendon 
(2225 N) must be equal to the force exerted on the foot by the tibia;
so
| | + | | = | 
|
so force exerted on the foot by the tibia will be;
| | = | | + | |
so we substitute IN OUR VALUES
| | = 750 N + 2225 N
| | = 2975 N
Therefore, the force exerted on the foot by the tibia would be 2975 N
