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
Answer:
f = 692 N
Explanation:
given data:
f =800N
a =1.2 m s^{2}
m= 90 kg
from newton's second law
net force 
therefore we have from above equation
ma =F - f
putting all value to get force of friction
1.2*90 = 800 - f
f = 692 N
Explanation:
Liquids also exert pressure in all directions on the walls of the container they are stored in. We see water coming out from leaking pipes and taps. ... Gases (Air) also exert pressure in all directions
Answer: 1896.55J/kg°C
Explanation:
The quantity of Heat Energy (Q) required to heat a material depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = 1320 joules
Mass of material = 5.61kg
C = ? (let unknown value be Z)
Φ = 0.124°C
Then, Q = MCΦ
1320J = 5.61kg x Z x 0.124°C
1320J = 0.696kg°C x Z
Z = (1320J / 0.696kg°C)
Z = 1896.55 J/kg°C
Thus, the specific heat of the material is 1896.55J/kg°C
D. Velocity because it describes a speed and direction
