Answer:
0.20kg-m^2
Explanation:
Let the linear velocity of the rope(=of pulley) is v m/s
Using kinematic equation
=> v = u + at
=>v = 0 + 4.9a
=>v = 4.9a ------------ eq1
By v^2 = u^2 + 2as
=>v^2 = 0 + 2 x v/4.9 x 1.2
=>4.9v^2 - 2.4v = 0
=>v(4.9v - 2.4) = 0
=>v = 2.4/4.9 = 0.49 m/s
Thus by v = r x omega
=>omega = v/r = 0.49/0.02 = 24.49 rad/sec
BY W = F x s = 50 x 1.2 = 60 J
=>KE(rotational) = W = 1/2 x I x omega^2
=>60 = 1/2 x I x (24.49)^2
=>I = 0.20 kg-m^2
a)., b)., and c). are completely false.
There isn't a grain of truth among them.
In Physics, the technical definition of 'Work' is (force) times (distance).
The suspended ash made for some some spectacular sunsets! Sulfuric acid was spread worldwide, increasing acidity of rain. Ash deflected energy from the sun, causing a slight drop on global temps for a few years.
Answer:
Out of this, Area is not a fundamental physical quantity.
Answer:
a = - 0.248 m/s²
Explanation:
Frictional drag force
F = ½ *(ρ* v² * A * α)
ρ = density of air , ρ = 1.295 kg/m^3
α = drag coef , α = 0.250
v = 100 km/h x 1000m / 3600s
v = 27.77 m/s
A = 2.20m^2
So replacing numeric in the initial equation
F = ½ (1.295kg/m^3)(27.77m/s)²(2.30m^2)(0.26)
F = 298.6 N
Now knowing the force can find the acceleration
a = - F / m
a = - 298.6 N / 1200 kg
a = - 0.248 m/s²
