Answer:
0.54454
104.00902 N
Explanation:
m = Mass of wheel = 100 kg
r = Radius = 0.52 m
t = Time taken = 6 seconds
= Final angular velocity
= Initial angular velocity
= Angular acceleration
Mass of inertia is given by

Angular acceleration is given by

Equation of rotational motion

The coefficient of friction is 0.54454
At r = 0.25 m

The force needed to stop the wheel is 104.00902 N
Answer:
R1 = 5.13 Ω
Explanation:
From Ohm's law,
V = IR............... Equation 1
Where V = Voltage, I = current, R = resistance.
From the question,
I = 2 A, R = R1
Substitute into equation 1
V = 2R1................ Equation 2
When a resistance of 2.2Ω is added in series with R1,
assuming the voltage source remain constant
R = 2.2+R1, and I = 1.4 A
V = 1.4(2.2+R1)................. Equation 3
Substitute the value of V into equation 3
2R1 = 1.4(2.2+R1)
2R1 = 3.08+1.4R1
2R1-1.4R1 = 3.08
0.6R1 = 3.08
R1 = 3.08/0.6
R1 = 5.13 Ω
Answer:
9 and 3 N
Explanation:
Forces in the same direction sum up to produce the resultant force;
One force subtract the other will give the resultant force when they are in opposite directions;
Lets say one direction is forwards and the opposite backwards;
We have one force, let's say force A, in the forwards direction and another force, force B, acting in the same (forwards) or opposite (backwards) direction;
If B is acting in the same direction, then the resultant force (in this case) will be as follows:
A + B = 12
If B is acting in the opposite direction, then the resultant force will be as follows:
A - B = 6
Summing the two equations will allow us to solve for A:
A + B + (A - B) = 12 + 6
2A = 18
A = 9
Substitute this into either of the above equations and we can solve for B:
(9) - B = 6
B = 9 - 6
B = 3
A closed circle means the number is included and an open circle means its not.
Answer:

Explanation:
Given that
Radius of track = R
Radius of ball = r
The ball can be treated as solid sphere, so
The moment of inertia of ball

When the ball reach at the lowest position then it will have both angular and linear speed.
Condition for rolling without slipping v= ωr
Form energy conservation

v= ωr





