<u>The answer is not contained detail explanation, just a solution and the required values. </u>
All the details are in the pictures, the answers are marked with orange colour.
Note,
in the task no 20.:
V - the velocity of the pair of the balls after collision.
in the task no 21:
m₁ - the mass of the copper ball; m₂ - the mass of the copper calorimeter; m₃ - the mass of the water; t₀ - the initial temperature of water in the copper calorimeter; θ - the final temperature in the calorimeter after the copper ball is transferred into a copper calorimeter; t₁ - the required initial temperature of the copper ball before it is transferred into the calorimeter.
Answer:
Explanation:
STEP 1
<u>Given</u>
Radius of cylinder = r = 25cm, 2.5m
mass = 27kg
cylinder is mounted so as to rotate freely about a horizontal axis that is parallel to and 60cm to the central logitudinal axis of the cylinder
height = 0.6m
<u>part 1</u>
The cylinder is mounted so as to rotate freely about a horizontal axis tha is paralle to 60cm from the central longitudinal axis of then cylinder. The rotational inertia of the cylinder about the axis of rotation is given by
<em>I = Icm + mh²</em>
<em>∴ I = 1/2mr² + mh² = 1/2x27x (0.5)² + 20 x (0.6)²</em>
<em>I=13.09kg.m²</em>
where
<em>I</em>cm is the rotational inertia of the cylinder about its central axis
m is the mass of the cylinder
h is the distance between the axis of the rotation and the central axis of the cylinder
r is the radius of the cylinder
<em> </em><em> I=13.09kg.m²</em>
<em>part2</em>
<em>from the conservation of the total mechanical energy of the meter stick, the change in gravitational potential energyof the meter stick plus the change in kinetic energy must be zero</em>
<em>Δk + Δu = 0</em>
<em>1/2 </em>I(w²-w²) = Ui-Uf
1/2 x 13.09w² = mgh
∴w=√20 x 9.8 x 0.6/(1/2 x 13.09) =117.6/6.5
w=18.09rad/s
Answer: The angle of resolution =47.05°
Explanation:
The angle of resolution represents the resolve power and precision of optical instruments such as the eye, camera and e en a microscope. It is given by the equation:
Sin A = 1.220(W÷D)
Sin A= 1.220(300÷500)
Sin A = 1.220(0.6)
Sin A = 0.732
A= Sin^-1 0.732
A= 47.05°
Answer:
a = 8 m/s^2, Ffriction = 10 N, μk = 0.205
Explanation:
a. Force = Mass*Acceleration,
(since you didn't add the units..."5 block"....for the mass, I will assume it to be in kg, per SI units)
40 N = 5 kg*acceleration,
a = 40/5 = 8 m/s^2
b. As you know newtons second law (F=m*a) is actually in the form Fnet = m*a. Which means that if the friction force comes into play, it would be Fapplied - Ffriction = m*a.
Fapplied - Ffriction = m*a,
40 - Ffriction = 5*6,
40 - Ffriction = 30,
Ffriction = 40 - 30 = 10 N
c. The coefficient of kinetic friction is calculated by the formula "Ffriction = μk*Fnormal".
10 = μk*Fnormal (Fnormal = m*g = 5*9.8)
10 = μk*49,
μk=10/49 ≈ 0.205
