Fill in the fraction: 3,600/90 = 40; turn it into a unit fraction.
40 mi/min
Answer:
magnitude of the frictional torque is 0.11 Nm
Explanation:
Moment of inertia I = 0.33 kg⋅m2
Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s
Final angular velocity w = 0 (since it stops)
Time t = 13 secs
Using w = w° + §t
Where § is angular acceleration
O = 4.34 + 13§
§ = -4.34/13 = -0.33 rad/s2
The negative sign implies it's a negative acceleration.
Frictional torque that brought it to rest must be equal to the original torque.
Torqu = I x §
T = 0.33 x 0.33 = 0.11 Nm
The velocity is 14 m/s
The parameters given on the question are
mass= 0.060 kg
kinetic energy= 5.9 joules
K.E= 1/2mv²
5.9= 1/2 × 0.060 × v²
5.9= 0.5 × 0.060v²
5.9= 003v²
v²= 5.9/0.03
v²= 196.66
v= √196.66
v= 14 m/s
Hence the velocity of the egg before it strikes the ground is 14 m/s
brainly.com/question/2084569?referrer=searchResults
I believe that you would weigh around 68 or 69 N, or 7 kilograms.
Answer:
The speed is 1.52 m.
Explanation:
Given that,
Displacement =0.270 m
Distance = 0.130 m
Suppose a 0.321-kg mass is attached to a spring with a force constant of 13.3 N/m.
We need to calculate the angular velocity
Using formula of angular velocity
Put the value into the formula
We need to calculate the velocity
Using formula of velocity
Put the value into the formula
Hence, The speed is 1.52 m.
