Answer:
40 meters. look for the dot above the 20 on the x-axis and follow it over to the left.
Explanation:
Answer:
the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Explanation:
Given the data in the question;
we make use of the following expression;
hall Voltage VH = IB / ned
where I = 2.25 A
B = 0.685 T
d = 0.107 mm = 0.107 × 10⁻³ m
e = 1.602×10⁻¹⁹ C
VH = 2.59 mV = 2.59 × 10⁻³ volt
n is the electron density
so from the form; VH = IB / ned
VHned = IB
n = IB / VHed
so we substitute
n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )
n = 1.54125 / 4.4396226 × 10⁻²⁶
n = 3.4716 × 10²⁵ m⁻³
Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Answer:
70.07 Hz
Explanation:
Since the sound is moving away from the observer then
and
when moving towards observer
With
of 76 then taking speed in air as 343 m/s we have


Similarly, with
of 65 we have

Now

v_s=27.76 m/s
Substituting the above into any of the first two equations then we obtain

Answer:
A) coil A
Explanation:
According to Faraday, Induced emf is given as;
E.M.F = ΔФ/t
ΔФ = BACosθ
where;
ΔФ is change in magnetic flux
θ is the angle between the magnetic field, B, and the normal to the loop of area A
A is the area of the loop
B is the magnetic field
From the equation above, induced emf depends on the strength of the magnetic field.
Both coils have the same area and are oriented at right angles to the field.
Coil A has a magnetic field strength of 10-T which is greater than 1 T of coil B, thus, coil A will have a greater emf induced in it.
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 ×
m
dynamic viscosity = 1.75 ×
Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ 
so
= µ
............1
put here value
= 1.75×
× 
= 0.175 v
and
area between air and puck is given by
Area =
area =
area = 7.85 ×
m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 × 
force = 1.374 ×
v
and now apply newton second law
force = mass × acceleration
- force = 
- 1.374 ×
v = 
t = 
time = 2.18
so time required after impact for a puck is 2.18 seconds