Answer:
Magnetic field, B = 0.199 T
Explanation:
It is given that,
Radius of circular loop, r = 11.7 cm = 0.117 m
Magnetic flux through the loop, 
The magnetic flux linked through the loop is :


Here, 

or


B = 0.199 T
So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.
Answer:
a)6.34 x
W/m²
b)1.37 x
W/m²
c) see explanation.
Explanation:
a)The relation of intensity'I' of the radiation and area 'A' is given by:
I= P/A
where P= power of sunlight i.e 3.9 x
J
and the area of the sun is given by,
A= 4π
=> 4π
A=6.15 x
m²
= 3.9 x
/ 6.15 x
=><u> 6.34 x </u>
<u>W/m²</u>
b) First determine the are of the sphere in order to determine intensity at the surface of the virtual space
A= 4π
Now R= 1.5 x
m
A= 4π x 1.5 x
=>2.83 x
m²
The power that each square meter of Earths surface receives
= 3.9 x
/2.83 x
=><u>1.37 x </u>
<u> W/m²</u>
<u />
c) in part (b), by assuming the shape of the wavefront of the light emitted by the Sun is a spherical shape so each point has the same distance from the source i.e sun on the wavefront.
Answer:
KE = 1/2 M V^2 = 1/2 * 25 * 10^2 = 1250 J
Check
M2 = 1/2 M1
V2 = V1 / 2
E2 = 1/2 * 1/4 E1 = E1 / 8 = 10000 / 8 = 1250 J
Answer:
23.5 mV
Explanation:
number of turn coil 'N' =22
radius 'r' =3.00 cm=>
0.03m
resistance = 1.00 Ω
B= 0.0100t + 0.0400t²
Time 't'= 4.60s
Note that Area'A' = πr²
The magnitude of induced EMF is given by,
lƩl =ΔφB/Δt = N (dB/dt)A
=N[d/dt (0.0100t + 0.0400 t²)A
=22(0.0100 + 0.0800(4.60))[π(0.03)²]
=0.0235
=23.5 mV
Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV
Denser materials tend to be closer to earths center due to their mass gravity is shown by the equation mg
Which stands for mass x gravity.