B. only its velocity should change
In simple words, flux can be stated as the rate of flow of a fluid, radiant energy, or particles across a given area.
<u>Explanation:</u>
<u>Mutual Flux:</u>
- The magnetic lines present in among two magnets or solenoid is mutual flux.
- These are the lines in which the attraction and repulsion happens.
- The SI unit of mutual flux is the Henry
<u>Leakage Flux:</u>
- In simple words, it can be stated as the magnetic flux which does not follow the specially designed way in a magnetic circuit.
- Leakage flux in the induction motor takes spot due to current runs through the essence of the induction motor.
- The SI unit of Leakage flux is the Weber
<u>Magnetizing flux</u>
- Magnetic flux is an analysis of the entire magnetic field which moves in a given field
- In simple words can be defined as the Magnetic flux is what generates the field around a magnetic material.
- The SI unit of magnetic flux is the Weber
Answer:
θ = 225 rad
Explanation:
given data
angle = 25 rad
to find out
angular velocity after 3t?
solution
let angular acceleration α in t
θ = ω × t + 0.5 × α × t² ........................1
here ω = 0 (initial velocity )
so put this value here
25 = 0 + 0.5 × α × t² ..........................2
α = 25 ÷ (0.5 t²)
α = 50 ÷ t² .........................3
now here we take in 3t
θ = ω × 3t + 0.5 × α × (3t)²
for ω = 0
θ = 0 + 0.5 × α × 9t²
now put value in eq 2
so
θ = (0.5) × (50 ÷ t²) × (3t)²
θ = 25 × 9
θ = 225 rad
Answer:
(a) -1.18 m/s
(b) 0.84 m/s
Explanation:
(a)
The total linear momentum before the lumberjack begins to move is zero because all parts of the system are at res
From the law of conservation of momentum
m1v1+m2v2=0 hence m1v1=-m2v2 where m1 is mass of lumberjack, v1 is velocity of lumberjeck, m2 is mass of floating log, v2 is velocity of the floating log.
Substituting M1 for 103 Kg, V1 for 2.93 m/s, M2 for 255 Kg into the above equation we obtain
103Kg*2.93 m/s=-255Kg*V2
V2=-(103 kg*2.93 m/s)/255=-1.183490196 m/s
Hence V2=-1.18 m/s
(b)
For the second log
V(M1+M2)=m1v1 where V is the common velocity
V(103 Kg+255 Kg)=103 Kg*2.93 m/s
V=(103 Kg*2.93 m/s)/(103 Kg+255 Kg)=0.842988827 m/s
V=0.84 m/s
Answer:
a

b

Explanation:
From the question we are told that
The spring constant is 
The maximum extension of the spring is 
The number of oscillation is 
The time taken is 
Generally the the angular speed of this oscillations is mathematically represented as

where T is the period which is mathematically represented as

substituting values


Thus


this angular speed can also be represented mathematically as

=> 
substituting values


In SHM (simple harmonic motion )the equation for velocity is mathematically represented as

The velocity is maximum when

=> 
=> 
=> 