Answer:
Mass of the aluminium chunk = 278.51 g
Explanation:
For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same
A specific heat formula is given as
Energy Change = Mass of liquid x Specific Heat Capacity x Change in temperature
Q = m×c×ΔT
Heat gain by aluminium + heat lost by copper = 0 (1)
For Aluminium:
Q = 
Q = m x 17.94 joule
For Copper:
Q= 4996.53 Joule
from eq 1
m x 17.94 = 4996.53
Mass of the aluminium chunk = 278.51 g
here in the given situation if monkey starts free fall at the same instant when veterinarian shoots towards it then we know that vertical component of motion of monkey and the dart will be same as under gravity
so here the dart will always hit the monkey because they both moves under same acceleration
so here for the angle we can use
now we have
H = 3 m
L = 87.5 m
now we will have
so angle will be 1.96 degree above the ground
Given Information:
Current = I = 2.5 A
Magnetic field = B = 0.10 T
Radius = r = d/2 = 0.02/2 = 0.01 m
Length = L = 8 cm = 0.08 m
Required Information:
Number of turns = N = ?
Answer:
Number of turns = N ≈ 2547 turns
Step-by-step explanation:
The approximate model to find the number of turns is given by
B = μ₀nI
Where n = N/L
so
B = μ₀NI/L
N = BL/μ₀I
Where B is the magnetic field, L is the length of the solenoid, I is the current and μ₀ is the permeability of free space
N = (0.10*0.08)/(4πx10⁻⁷*2.5)
N ≈ 2547 Turns
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
