<span>Δ</span>E = q + w
q = heat (quantity of)
q and w can be positive or negative depending on if work/heat is being absorbed/done on the system or released/done by the system
Answer:
0.11 kg
Explanation:
Ft = MV
Ft = momentum 5.22kg m/s
M = mass
V = velocity 48.3m/s
Therefore
5.22 = M x 48.3
Divide both sides by 48.3
5.22/48.3 = M x 48.3/48.3
0.11 = M
M = 0.11kg
Answer:
Option A is correct.
(The faster object encounters more resistance)
Explanation:
Option A is correct. (The faster object encounters more resistance)
Air resistance depends on various factors:
- Speed of the object
- Cross-sectional area of the object
- Shape of the object
Formula:

As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.
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