Answer/solution:
Given :
Mass =5kg
T 1 =20 C,T 2 =100 ∘C
ΔT=100−20=80 ∘C
Q=m×C×ΔT
where C= specific heat capacity of water
=4200J/(kgK)
Q=5×4200×80
=1680000 Joule.
=1680KJ
Answer:
The West is known for "wide, open spaces", cattle, mines, and mountains.
Answer:
(a) 30 m/sec
(b) -50 m/sec
Explanation:
We have given initial velocity of ball u = 50 m/sec
Acceleration due to gravity 
(a) Time t = 2 sec
Now according to first equation of v = u-gt
So v=50-10×2=30 m/sec
(b) Time t = 10 sec
Now according to first equation of motion
So final velocity v = u-gt = 50-10×10 =-50 m/sec
Answer:
a. 0.342 kg-m² b. 2.0728 kg-m²
Explanation:
a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m
I = 1/2MR²
= 1/2 × 56.5 kg × (0.11 m)²
= 0.342 kgm²
So the moment of inertia of the skater is
b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)
I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m
I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²
= 0.1802 kg-m² + 0.6852 kg-m²
= 0.8654 kg-m²
The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².
So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²
Since f=ma when f=4.0 and m=2.0 so the magnitude of acceleration for the block is 2.0m
