Answer:
the net toque is τ=8.03* 10⁻⁴ N*m
Explanation:
Assuming the disk has constant density ρ, the moment of inertia I of is
I = ∫r² dm
since m = ρ*V = ρπR² h , then dm= 2ρπh r dr
thus
I = ∫r²dm = ∫r²2ρπh r dr =2ρπh ∫r³ dr = 2ρπh (R⁴/4- 0⁴/4)= ρπhR⁴ /2= mR²/2
replacing values
I = mR²/2= 0.017 kg * (0.06 m)²/2 = 3.06 *10⁻⁵ kg*m²
from Newton's second law applied to rotational motion
τ= Iα , where τ=net torque and α= angular acceleration
since the angular velocity ω is related with the angular acceleration through
ω= ωo + α*t → α =(ω-ωo)/t = (21 rad/s-0)/0.8 s = 26.25 rad/s²
therefore
τ= Iα= 3.06 *10⁻⁵ kg*m²*26.25 rad/s² = 8.03* 10⁻⁴ N*m
<span>Each laid 250 bricks but while Jake was still working, Josh was lounging in the shade. Josh has more power but that power was only on for 3 hours out of 4.5. Obviously Josh could get more done is less time as long as he keeps working. Jake will get the hang of it soon.</span>
Explanation:
It is given that,
Velocity of the electron, 
Magnetic field, 
Charge of electron, 
(a) Let
is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

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(b) The charge of electron, 
The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.
Answer:
If one plays ice hockey he should must have necessary equipment to support his sport. These equipment include Ice Skates,Helmet with Cage and Mouth-guard:, Hockey stick, Hockey pants,Hockey gloves, shoulder pads, elbow pads, Shin Guard:Neck guard and Jockstrap (men) or Pelvic protector (women)
Explanation:
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.