The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.
For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.
As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.
i.e.
P.E = K.E + R.K.E
Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
Learn more about angular velocity here:
brainly.com/question/1452612
C because it’s not a or B so 50/50 c or d and d is def not the answer so c
Electrons that are further away from the nucleus have more energy. As they enter an "excited" state, they jump up orbits.
Answer:
Given:
Resistance (R) = 100 Ω
Current (I) = 5 A
Time (t) = 2 hours
To Find:
Heat developed (H) in the electric iron
Explanation:
Formula:
Substituting values of I, R & t in the equation:
Heat developed (H) in the electric iron = 15 kWh
Diameter = 60 cm, Radius = 60/2 = 30 cm = 30/100 = 0.3 m.
The pebble in the tread goes by 3 times every second.
This is the same as 3 times per second.
Recall the unit of frequency is Hertz or per second, s⁻¹
So 3 times per second, Frequency, f = 3s⁻¹ or 3 Hertz
For angular motion:
Angular speed, ω = 2πf
= 2*π*3
= 6π rad/s
Linear speed, v = ωr = 6π * 0.3 = 1.8π m/s
Linear acceleration, a = v² / r
a = 1.8π * 1.8π / 0.3 = 10.8π² m/s²
Angular acceleration α = a/r = 10.8π² / 0.3 = 36π² rad/s²
Angular speed = 6π rad/s ≈ 18.840 rad/s
The linear speed of the pebble = 1.8π m/s ≈ 5.655 m/s
The angular acceleration = 36π² rad/s² ≈ 355.306 rad/s²
The linear acceleration of the pebble = 10.8π² m/s ≈ 106.592 m/s²
