If a space probe sends back info on an object traveling through space at close to the speed of light astronomers will be working
1 answer:
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Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Answer:
Mass of the aluminium chunk = 278.51 g
Explanation:
For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same
A specific heat formula is given as
Energy Change = Mass of liquid x Specific Heat Capacity x Change in temperature
Q = m×c×ΔT
Heat gain by aluminium + heat lost by copper = 0 (1)
For Aluminium:
Q =
Q = m x 17.94 joule
For Copper:
Q= 4996.53 Joule
from eq 1
m x 17.94 = 4996.53
Mass of the aluminium chunk = 278.51 g