•THAT THE PROPAGATION OF SOUND WAVES NEED MEDIUM TO TRAVEL
•THE MEDIUM SHOULD POSSES ELASTICITY
•FOR THE FASTER PROPAGATION OF SOUND THE PARTICLES SHOULD BE VERY CLOSE TO EACH OTHER
The mechanical efficiency = actual work / ideal work
So ζ = 1540 / 1600 * 100% = 96.25%
Answer:
a) 1.3 rad/s
b) 0.722 s
Explanation:
Given
Initial velocity, ω = 0 rad/s
Angular acceleration of the wheel, α = 1.8 rad/s²
using equations of angular motion, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
where
θ2 - θ1 = 53.2 rad
t2 - t1 = 7s
substituting these in the equation, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²
53.2 = 7.ω(0) + 1/2 * 1.8 * 49
53.2 = 7.ω(0) + 44.1
7.ω(0) = 53.2 - 44.1
ω(0) = 9.1 / 7
ω(0) = 1.3 rad/s
Using another of the equations of angular motion, we have
ω(0) = ω(i) + α*t1
1.3 = 0 + 1.8 * t1
1.3 = 1.8 * t1
t1 = 1.3/1.8
t1 = 0.722 s
Answer:
109656.25 Nm
Explanation:
= Final angular velocity = 1.5 rad/s
= Initial angular velocity = 0
= Angular acceleration
t = Time taken = 6 s
m = Mass of disk = 29000 kg
r = Radius = 5.5 m

Torque is given by

The torque specifications must be 109656.25 Nm
Answer:
electrons
Explanation:
The magnitude of the electric field outside an electrically charged sphere is given by the equation

where
k is the Coulomb's constant
Q is the charge stored on the sphere
r is the distance (from the centre of the sphere) at which the field is calculated
In this problem, the cloud is assumed to be a charged sphere, so we have:
is the maximum electric field strength tolerated by the air before breakdown occurs
is the radius of the sphere
Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Assuming that the cloud is negatively charged, then

And since the charge of one electron is

The number of excess electrons on the cloud is
