Answer:
θ = 56.09°
Explanation:
Given that
h = 8.5 ft
speed ,u= 28 ft/s
We know that acceleration due to gravity g
g= 32.2 ft/s²
The maximum height h is given as
![h=\dfrac{u^2sin^2\theta}{2g}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7Bu%5E2sin%5E2%5Ctheta%7D%7B2g%7D)
Now by putting the values
![8.5=\dfrac{28^2sin^2\theta}{2\times 32.2}](https://tex.z-dn.net/?f=8.5%3D%5Cdfrac%7B28%5E2sin%5E2%5Ctheta%7D%7B2%5Ctimes%2032.2%7D)
![sin^2\theta=0.69](https://tex.z-dn.net/?f=sin%5E2%5Ctheta%3D0.69%20)
![sin\theta=0.83](https://tex.z-dn.net/?f=sin%5Ctheta%3D0.83%20)
θ = 56.09°
Therefore the angle will be θ = 56.09°
Force = mass x acceleration
force = 2500kg x (20m/s / 10m/s)
force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN
i hope this is right (^^)
Answer:
linear acceleration
![a = \frac{2g}{2 + \frac{R}{b}}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B2g%7D%7B2%20%2B%20%5Cfrac%7BR%7D%7Bb%7D%7D)
angular acceleration
![\alpha = \frac{2g}{R(2 + \frac{R}{b})}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B2g%7D%7BR%282%20%2B%20%5Cfrac%7BR%7D%7Bb%7D%29%7D)
Explanation:
As we know that the force due to tension force is upwards while weight of the disc is downwards
so we will have
![2mg - T = 2ma](https://tex.z-dn.net/?f=2mg%20-%20T%20%3D%202ma)
also we have
![Tb = (\frac{1}{2}mR^2 + \frac{1}{2}mR^2)\alpha](https://tex.z-dn.net/?f=Tb%20%3D%20%28%5Cfrac%7B1%7D%7B2%7DmR%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DmR%5E2%29%5Calpha)
now we have
![Tb = mR^2(\frac{a}{R})](https://tex.z-dn.net/?f=Tb%20%3D%20mR%5E2%28%5Cfrac%7Ba%7D%7BR%7D%29)
![T = \frac{mRa}{b}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7BmRa%7D%7Bb%7D)
now we have
![2mg = (2ma + \frac{mRa}{b})](https://tex.z-dn.net/?f=2mg%20%3D%20%282ma%20%2B%20%5Cfrac%7BmRa%7D%7Bb%7D%29)
![a(2 + \frac{R}{b}) = 2g](https://tex.z-dn.net/?f=a%282%20%2B%20%5Cfrac%7BR%7D%7Bb%7D%29%20%3D%202g)
so we have
linear acceleration
![a = \frac{2g}{2 + \frac{R}{b}}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B2g%7D%7B2%20%2B%20%5Cfrac%7BR%7D%7Bb%7D%7D)
angular acceleration
![\alpha = \frac{2g}{R(2 + \frac{R}{b})}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B2g%7D%7BR%282%20%2B%20%5Cfrac%7BR%7D%7Bb%7D%29%7D)
Answer:
B
Explanation:
The kinetic energy in an object is converted into potential energy. This makes the kinetic decrease, while the potential increases.
Answer:
9.6 rad/s
Explanation:
= length of the metal rod = 50 cm = 0.50 m
= Mass of the long metal rod = 780 g = 0.780 kg
Moment of inertia of the rod about one end is given as
![I = \frac{ML^{2}}{3} = \frac{(0.780)(0.50)^{2}}{3} = 0.065 kgm^{2}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BML%5E%7B2%7D%7D%7B3%7D%20%3D%20%5Cfrac%7B%280.780%29%280.50%29%5E%7B2%7D%7D%7B3%7D%20%3D%200.065%20kgm%5E%7B2%7D)
= force applied by the hammer blow = 1000 N
Torque produced due to the hammer blow is given as
![\tau = \frac{FL}{2}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cfrac%7BFL%7D%7B2%7D)
![\tau = \frac{(1000)(0.50)}{2}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cfrac%7B%281000%29%280.50%29%7D%7B2%7D)
![\tau = 250 Nm](https://tex.z-dn.net/?f=%5Ctau%20%3D%20250%20Nm)
= time of blow = 2.5 ms = 0.0025 s
= Angular velocity after the blow
Using Impulse-change in angular momentum, we have
![I w = \tau t\\(0.065) w = (250) (0.0025)\\w = 9.6 rads^{-1}](https://tex.z-dn.net/?f=I%20w%20%3D%20%5Ctau%20t%5C%5C%280.065%29%20w%20%3D%20%28250%29%20%280.0025%29%5C%5Cw%20%3D%209.6%20rads%5E%7B-1%7D)