Answer:
A) 138.8g
B)73.97 cm/s
Explanation:
K = 15.5 Kn/m
A = 7 cm
N = 37 oscillations
tn = 20 seconds
A) In harmonic motion, we know that;
ω² = k/m and m = k/ω²
Also, angular frequency (ω) = 2π/T
Now, T is the time it takes to complete one oscillation.
So from the question, we can calculate T as;
T = 22/37.
Thus ;
ω = 2π/(22/37) = 10.5672
So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g
B) In simple harmonic motion, velocity is given as;
v(t) = vmax Sin (ωt + Φ)
It is from the derivative of;
v(t) = -Aω Sin (ωt + Φ)
So comparing the two equations of v(t), we can see that ;
vmax = Aω
Vmax = 7 x 10.5672 = 73.97 cm/s
Answer:
150J
Explanation:
Formula : <u>Work</u><u> </u><u>done</u>
Force x distance
work done = force x distance
Distance should be measured in meters
300÷100=3m
work done = 450 x 3
=150J
Answer:
-0.0047 rad/s²
335.103 seconds
99.18 seconds
Explanation:
= Final angular velocity
= Initial angular velocity = 1.5 ra/s
= Angular acceleration
= Angle of rotation = 40 rev
t = Time taken
Equation of rotational motion
![\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2](https://tex.z-dn.net/?f=%5Comega_f%5E2-%5Comega_i%5E2%3D2%5Calpha%20%5Ctheta%5C%5C%5CRightarrow%20%5Calpha%3D%5Cfrac%7B%5Comega_f%5E2-%5Comega_i%5E2%7D%7B2%5Ctheta%7D%5C%5C%5CRightarrow%20%5Calpha%3D%5Cfrac%7B0%5E2-1.5%5E2%7D%7B2%5Ctimes%202%5Cpi%20%5Ctimes%2040%7D%5C%5C%5CRightarrow%20%5Calpha%3D-0.0047%5C%20rad%2Fs%5E2)
Acceleration while slowing down is -0.0047 rad/s²
![t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Comega_f-%5Comega_i%7D%7B%5Calpha%7D%5C%5C%5CRightarrow%20t%3D%5Cfrac%7B0-1.5%7D%7B-0.0047%7D%5C%5C%5CRightarrow%20t%3D335.103%5C%20s)
Time taken to slow down is 335.103 seconds
![\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0](https://tex.z-dn.net/?f=%5Ctheta%3D%5Comega_it%2B%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2%5C%5C%5CRightarrow%2020%5Ctimes%202%5Cpi%3D1.5%5Ctimes%20t%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20-0.0047%5Ctimes%20t%5E2%5C%5C%5CRightarrow%200.00235t%5E2-1.5t%2B125.66%3D0)
Solving the equation
![t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-%5Cleft%28-1.5%5Cright%29%2B%5Csqrt%7B%5Cleft%28-1.5%5Cright%29%5E2-4%5Ccdot%20%5C%3A0.00235%5Ccdot%20%5C%3A125.66%7D%7D%7B2%5Ccdot%20%5C%3A0.00235%7D%2C%20%5Cfrac%7B-%5Cleft%28-1.5%5Cright%29-%5Csqrt%7B%5Cleft%28-1.5%5Cright%29%5E2-4%5Ccdot%20%5C%3A0.00235%5Ccdot%20%5C%3A125.66%7D%7D%7B2%5Ccdot%20%5C%3A0.00235%7D%5C%5C%5CRightarrow%20t%3D539.11%2C%2099.18%5C%20s)
The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103
A. 20m/s because the unit for velocity is m/s
Answer:
Grounded used to carry the current in the normal conditions. while on the other side grounding is refer to that safety wire that is connected to the earth and currently does not transmit through it.
As grounding wire is referred to as safety wire hence it carries current only in an emergency like a short circuit.
Explanation:
A grounded conductor is referred to as one of the wire that needed in an electric circuit. it is basically a neutral conductor. It used to carry the current in the normal conditions. while on the other side grounding is refer to that safety wire that is connected to the earth and currently does not transmit through it.
As grounding wire is referred to as safety wire hence it carries current only in an emergency like a short circuit.