In a completely inelastic collision, the two objects stick together after the collision.
Answer:
the moment of inertia of the wheel is 0.408 kg.m²
Explanation:
Given;
tangential force applied to the wheel, f = 90 N
radius of the wheel, r = 0.15 m
initial angular speed of the wheel, ω₁ = 0
final angular speed of the wheel, ω₂ = 14.3 rev/s
time of motion of the wheel, t = 2.72 s
The tangential acceleration of the wheel is calculated as;
![\alpha _t = \alpha r](https://tex.z-dn.net/?f=%5Calpha%20_t%20%3D%20%5Calpha%20r)
where;
is the angular acceleration
![\alpha = \frac{\Delta \omega }{\Delta t} = \frac{\omega _2 - \omega_1}{t_2-t_1}\\\\where;\\\\\omega_2 \ is \ the \ final \ angular \ speed = 14.3 \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 89.86 \ rad/s](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5CDelta%20%5Comega%20%7D%7B%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5Comega%20_2%20-%20%5Comega_1%7D%7Bt_2-t_1%7D%5C%5C%5C%5Cwhere%3B%5C%5C%5C%5C%5Comega_2%20%5C%20is%20%5C%20the%20%5C%20final%20%5C%20angular%20%5C%20speed%20%3D%2014.3%20%5Cfrac%7Brev%7D%7Bs%7D%20%5Ctimes%20%5Cfrac%7B2%5Cpi%20%5C%20rad%7D%7B1%20%5C%20rev%7D%20%3D%2089.86%20%5C%20rad%2Fs)
![\alpha = \frac{89.86 - 0}{2.72} = 33.04 \ rad/s^2\\\\\alpha _t = \alpha r\\\\\alpha _t = 33.04 \ rad/s^2 \times 0.15 \ m\\\\\alpha _t = 4.96 \ m/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B89.86%20-%200%7D%7B2.72%7D%20%3D%2033.04%20%5C%20rad%2Fs%5E2%5C%5C%5C%5C%5Calpha%20_t%20%3D%20%5Calpha%20r%5C%5C%5C%5C%5Calpha%20_t%20%3D%2033.04%20%5C%20rad%2Fs%5E2%20%5Ctimes%200.15%20%5C%20m%5C%5C%5C%5C%5Calpha%20_t%20%3D%204.96%20%5C%20m%2Fs%5E2)
The mass of the wheel is calculated as;
F = ma
m = F/a
m = (90)/(4.96)
m = 18.15 kg
The moment of inertia of the wheel is calculated as;
I = mr²
I = 18.15 x (0.15)²
I = 0.408 kg.m²
Therefore, the moment of inertia of the wheel is 0.408 kg.m²
Answer:1.301 s
Explanation:
Given
Initial Velocity(u)=30 m/s
Height of cliff=8.3 m
Time taken to cover 8.3 m
![h=ut+\frac{at^2}{2}](https://tex.z-dn.net/?f=h%3Dut%2B%5Cfrac%7Bat%5E2%7D%7B2%7D)
here Initial vertical velocity is 0
![8.3=\frac{gt^2}{2}](https://tex.z-dn.net/?f=8.3%3D%5Cfrac%7Bgt%5E2%7D%7B2%7D)
![t^2=1.69](https://tex.z-dn.net/?f=t%5E2%3D1.69)
![t=1.301 s](https://tex.z-dn.net/?f=t%3D1.301%20s)
Horizontal distance
![R=u\times t](https://tex.z-dn.net/?f=R%3Du%5Ctimes%20t)
![R=30\times 1.301=39.04 m](https://tex.z-dn.net/?f=R%3D30%5Ctimes%201.301%3D39.04%20m)
Answer:
the kinetic energy of clown A is 0.444 times the kinetic energy of clown B.
Explanation:
Let the spring constant of the spring is k.
For clown A:
m = 40 kg
let the extension in the spring is y.
So, the spring force, F = k y
m g = k y
40 x g = k x y
y = 40 x g / k ..... (1)
For clown B:
m' = 60 kg
Let the extension in the spring is y'.
So, the spring force, F' = k y'
m' g = k y'
y' = 60 x g / k .....(2)
Kinetic energy for A, K = 1/2 ky^2
Kinetic energy for B, K' = 1/2 ky'^2
So, K/K' = y^2/y'^2 K / K' = (40 x 40) / (60 x 60) (from equation (1) and (2))
K / K' = 0.444
K = 0.444 K'
So the kinetic energy of clown A is 0.444 times the kinetic energy of clown B.
Answer:
B = 32.17 x 10^-8 Tesla
u = 8.24 x 10^-8 J/m^3
P/A = 24.72 W/m^2
Explanation:
E = 96.5 V/m
velocity of light, c = 3 x 10^8 m/s
Let B be the magnetic field.
The relation between the electric field strength and the magnetic field strength is given by
B = E / c = 96.5 / (3 x 10^8) = 32.17 x 10^-8 Tesla
Let u be the energy density.
![u = \frac{1}{2}\times \varepsilon _{0}E^{2}+\frac{1}{2\mu _{0}}B^{2}](https://tex.z-dn.net/?f=u%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cvarepsilon%20_%7B0%7DE%5E%7B2%7D%2B%5Cfrac%7B1%7D%7B2%5Cmu%20_%7B0%7D%7DB%5E%7B2%7D)
![u = \frac{1}{2}\times 8.854\times 10^{-12}\times 96.5\times 96.5+\frac{1}{2\times 4\times 3.14\times 10^{-7}}\times 32.17^{2}\times 10^{-16}](https://tex.z-dn.net/?f=u%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%208.854%5Ctimes%2010%5E%7B-12%7D%5Ctimes%2096.5%5Ctimes%2096.5%2B%5Cfrac%7B1%7D%7B2%5Ctimes%204%5Ctimes%203.14%5Ctimes%2010%5E%7B-7%7D%7D%5Ctimes%2032.17%5E%7B2%7D%5Ctimes%2010%5E%7B-16%7D)
u = 8.24 x 10^-8 J/m^3
Let Power flow per unit area is
P/A = u x c = 8.24 x 10^-8 x 3 x 10^8 = 24.72 W/m^2