Answer:
a) 
, b) 
Explanation:
a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:
In this case, the force is equal to the weight of the object:
The spring constant is:
b) The length of the spring is:
Answer:
Scientific definitions for ferromagnetic
The property of being strongly attracted to either pole of a magnet. Ferromagnetic materials, such as iron, contain unpaired electrons, each with a small magnetic field of its own, that align readily with each other in response to an external magnetic field.
Explanation:
Answer:0.0704 kg
Explanation:
Given
initial Absolute pressure
=210+101.325=311.325
as the volume remains constant therefore
therefore Gauge pressure is 337.44-101.325=236.117 KPa
Initial mass 
Final mass 
Therefore 
=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back
Answer:
Explanation:
Given a square side loop of length 10cm
L=10cm=0.1m
Then, Area=L²
Area=0.1²
Area=0.01m²
Given that, frequency=60Hz
And magnetic field B=0.8T
a. Flux Φ
Flux is given as
Φ=BA Sin(wt)
w=2πf
Φ=BA Sin(2πft)
Φ=0.8×0.01 Sin(2×π×60t)
Φ=0.008Sin(120πt) Weber
b. EMF in loop
Emf is given as
EMF= -N dΦ/dt
Where N is number of turns
Φ=0.008Sin(120πt)
dΦ/dt= 0.008×120Cos(120πt)
dΦ/dt= 0.96Cos(120πt)
Emf=-NdΦ/dt
Emf=-0.96NCos(120πt). Volts
c. Current induced for a resistance of 1ohms
From ohms law, V=iR
Therefore, Emf=iR
i=EMF/R
i=-0.96NCos(120πt) / 1
i=-0.96NCos(120πt) Ampere
d. Power delivered to the loop
Power is given as
P=IV
P=-0.96NCos(120πt)•-0.96NCos(120πt)
P=0.92N²Cos²(120πt) Watt
e. Torque
Torque is given as
τ=iL²B
τ=-0.96NCos(120πt)•0.1²×0.8
τ=-0.00768NCos(120πt) Nm
Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed.
P=F/A
