Answer:
The approximate change in entropy is -14.72 J/K.
Explanation:
Given that,
Temperature = 22°C
Internal energy 
Final temperature = 16°C
We need to calculate the approximate change in entropy
Using formula of the entropy

Where,  = internal energy
 = internal energy
T = average temperature
Put the value in to the formula


Hence, The approximate change in entropy is -14.72 J/K.
 
        
             
        
        
        
Answer:
1.06 secs
Explanation:
Initial speed of sled, u = 8.4 m/s
Final speed of sled, v = 5.8 m/s
Coefficient of kinetic friction, μ = 0.25
Using the impulse momentum theory, we know that the impulse applied to the sled is equal to change in momentum of the sled:
 FΔt = mv - mu
where m = mass of the object
Δt = time interval
F = force applied 
The force applied on the sled is the frictional force, which is given as:
F = -μmg
where g = acceleration due to gravity
Therefore:
-μmgΔt =  mv - mu
-μmgΔt = m(v - u)
-μgΔt = v - u
Making Δt subject of formula:
Δt = (v - u) / -μg
Δt = (5.8 - 8.4) / (-0.25 * 9.8)
Δt = -2.6/ -2.45
Δt = 1.06 secs
It took the sled 1.06 secs to travel from A to B.
 
        
             
        
        
        
Answer:

Explanation:
Total spectators = 5000
Counted by the groups of ten, So at last the result will be:
=> 5000/10 = 500
Significant figures in 500 are 3
 
        
             
        
        
        
Answer:
A) μ = A.m² 
B) z = 0.46m 
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil 
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So, 
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m = 
 
        
             
        
        
        
This is an example of resonance - when one object vibrating at the same natural frequency of a second object forces that second object into vibrational motion. The result of resonance is always a large vibration.
Answer D. Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body Because this statement contridicts the above statement, it is not accurate