The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
By definition, the gain in PE (potential energy) is
ΔPE = m*g*h
Given:
mg = 40 N (Note that m*g = weight)
h = 5 m
ΔPE = (40 N)*(5 m) =200 J
Answer: 200 J
Answer:
1.71 km
Explanation:
Convert 30 minutes to seconds:
30 min × (60 s / min) = 1800 s
Find the displacement:
0.95 m/s × 1800 s = 1710 m
Convert to kilometers:
1710 m × (1 km / 1000 m) = 1.71 km
Answer:
Explanation:
Let Torque due to friction be
F
Net torque
= 46 - F
Angular impulse = change in angular momentum
=( 46 - F ) x 17 = I X 580
When external torque is removed , only friction creates torque reducing its speed to zero in 120 s so
Angular impulse = change in angular momentum
F x 120 = I X 580
( 46 - F ) x 17 = F x 120
137 F = 46 x 17
F = 5.7 Nm
b )
Putting this value in first equation
5.7 x 120 = I x 580
I = 1.18 kg m²
Answer:
2.59 Kg, 3.25 Kg
Explanation:
Acceleration can be found using equation
where s is the release distance, a is acceleration and t is time
Making a the subject of the formula
Substituting 0.28 for s and time for 1 second
Acceleration formula for the Atwood machine is given by
where m1 and m2 are first and second masses respectively
Two situations are possible
When m1>m2
Assuming m1 is 3.7kg which is heavier than m2
Substituting a for 
and m1 as 2.9Kg and taking acceleration due to gravity as 
10.37m2=28.449-1.624
10.37m2=26.825
m2=2.59 Kg
<u>When m1<m2</u>
Then m1=2.9Kg hence
-9.25m2=-28.449-1.624=-30.073
9.25m2=30.073
m2=3.25 Kg
