Answer:
![313.92\ \text{m/s}](https://tex.z-dn.net/?f=313.92%5C%20%5Ctext%7Bm%2Fs%7D)
![47.088\ \text{kg m/s}](https://tex.z-dn.net/?f=47.088%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
Explanation:
m = Mass of ball = 150 g
= Angle of kick = ![60^{\circ}](https://tex.z-dn.net/?f=60%5E%7B%5Ccirc%7D)
= Displacement of ball in x direction = 12 m
Range of projectile is given by
![x=\dfrac{u\sin^2\theta}{2g}\\\Rightarrow u=\dfrac{2xg}{\sin^2\theta}\\\Rightarrow u=\dfrac{2\times 12\times 9.81}{\sin^260^{\circ}}\\\Rightarrow u=313.92\ \text{m/s}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7Bu%5Csin%5E2%5Ctheta%7D%7B2g%7D%5C%5C%5CRightarrow%20u%3D%5Cdfrac%7B2xg%7D%7B%5Csin%5E2%5Ctheta%7D%5C%5C%5CRightarrow%20u%3D%5Cdfrac%7B2%5Ctimes%2012%5Ctimes%209.81%7D%7B%5Csin%5E260%5E%7B%5Ccirc%7D%7D%5C%5C%5CRightarrow%20u%3D313.92%5C%20%5Ctext%7Bm%2Fs%7D)
The velocity of the ball instantly after the man kicks the ball is ![313.92\ \text{m/s}](https://tex.z-dn.net/?f=313.92%5C%20%5Ctext%7Bm%2Fs%7D)
Impulse is given by
![J=mu\\\Rightarrow J=0.15\times 313.92\\\Rightarrow J=47.088\ \text{kg m/s}](https://tex.z-dn.net/?f=J%3Dmu%5C%5C%5CRightarrow%20J%3D0.15%5Ctimes%20313.92%5C%5C%5CRightarrow%20J%3D47.088%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
The impulse of his foot on the ball is
.
Explanation:
Given that,
2 strings both vibrate at exactly 220 Hz. The frequency of sound wave depends on the tension in the strings.
The tension in one of them is then decreased sightly, then
will decrese.
Beat frequency, ![f=f_1-f_2](https://tex.z-dn.net/?f=f%3Df_1-f_2)
![3=220-f_2](https://tex.z-dn.net/?f=3%3D220-f_2)
So, the new frequency of the string is 217 Hz. Hence, this is the required solution.
Answer:
The average angular acceleration is -2.628 rad/s²
Explanation:
Counterclockwise = positive
Clockwise = -negative
Given;
initial rotation of the flywheel, θ₁ = 6.55 rotation/s
final rotation of the flywheel, θ₂ = - 2.19 rotation/s
The average angular acceleration is given by;
![\alpha = \frac{\delta \theta}{\delta t}\\\\ \alpha =\frac{\theta _2 - \theta_ 1}{t}\\\\ \alpha =\frac{-2.19 -6.55}{20.9} \\\\ \alpha =\frac{-8.74}{20.9}\\\\ \alpha = -0.4182 \ rotation / s^2\\\\ \alpha = \frac{-0.4182 \ rotation}{s^2}*\frac{2\pi \ radian}{rotation}\\\\ \alpha = -2.628 \ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Cdelta%20%5Ctheta%7D%7B%5Cdelta%20t%7D%5C%5C%5C%5C%20%5Calpha%20%3D%5Cfrac%7B%5Ctheta%20_2%20-%20%5Ctheta_%201%7D%7Bt%7D%5C%5C%5C%5C%20%5Calpha%20%3D%5Cfrac%7B-2.19%20-6.55%7D%7B20.9%7D%20%5C%5C%5C%5C%20%5Calpha%20%3D%5Cfrac%7B-8.74%7D%7B20.9%7D%5C%5C%5C%5C%20%5Calpha%20%3D%20-0.4182%20%5C%20rotation%20%2F%20s%5E2%5C%5C%5C%5C%20%5Calpha%20%3D%20%5Cfrac%7B-0.4182%20%5C%20rotation%7D%7Bs%5E2%7D%2A%5Cfrac%7B2%5Cpi%20%5C%20radian%7D%7Brotation%7D%5C%5C%5C%5C%20%5Calpha%20%3D%20-2.628%20%5C%20rad%2Fs%5E2)
Therefore, the average angular acceleration is -2.628 rad/s²
Answer:
can you show a graph but if not i believe the answer is x=6m
Explanation: