<span>
The needle of a compass will always lies along the magnetic
field lines of the earth.
A magnetic declination at a point on the earth’s surface
equal to zero implies that
the horizontal component of the earth’s magnetic field line
at that specific point lies along
the line of the north-south magnetic poles. </span>
The presence of a
current-carrying wire creates an additional <span>
magnetic field that combines with the earth’s magnetic field.
Since magnetic
<span>fields are vector quantities, therefore the magnetic field of
the earth and the magnetic field of the vertical wire must be
combined vectorially. </span></span>
<span>
Where:</span>
B1 = magnetic field of
the earth along the x-axis = 0.45 × 10 ⁻ ⁴ T
B2 = magnetic field due to
the straight vertical wire along the y-axis
We can calculate for B2
using Amperes Law:
B2 = μ₀ i / [ 2 π R ]
B2 = [ 4π × 10 ⁻ ⁷ T • m / A ] ( 36 A ) / [ 2 π (0.21 m ) ] <span>
B2 = 5.97 × 10 ⁻ ⁵ T = 0.60 × 10 ⁻ ⁴ T </span>
The angle can be
calculated using tan function:<span>
tan θ = y / x = B₂ / B₁ = 0.60 × 10 ⁻ ⁴ T / 0.45 × 10 ⁻ ⁴ T <span>
tan θ = 1.326</span></span>
θ = 53°
<span>
<span>The compass needle points along the direction of 53° west of
north.</span></span>
Answer:
C. Talk test
Explanation:
The talk test would be readily available to me at the the least cost.
The talk test is about the easiest way that one can monitor intensity as they exercise. Because the only thing needed here is the ability to talk and to breathe.
The intensity lies on if one can talk and breathe at the same time. The harder one exercises, the more breathless they become and they find it difficult to talk.
Answer:
0.58 days is needed
Explanation:
Gravitational energy change from a height change equal to 8848 m for a mass of 53 kg is:
m*g*h = 53*9.81*8848 = 4600340.64 J = 4600.34 kJ
4.184 kJ are equivalent to 1 kcal, then:
4.184 kJ / 4600.34 kJ = 1 kcal / x kcal
x = 4600.34/4.184
x = 1099.5 kcal or large calories
The diet is 1910 kcal/day, then you need 1099.5/1910 = 0.58 days
Complete Question
The light energy that falls on a square meter of ground over the course of a typical sunny day is about 20 MJ . The average rate of electric energy consumption in one house is 1.0 kW .
If light energy to electric energy conversion using solar cells is 12 % efficient, how many square miles of land must be covered with solar cells to supply the electrical energy for 350000 houses? Assume there is no cloud cover.
Answer:
The area is 
Explanation:
From the question we are told that
The efficiency is 
12%
The number of houses is 
The light energy per day is 
The average rating of electric energy for a house is 
Generally the electric energy which the solar cells covering 
produces in a day is
Energy for required by one house for one day is
Energy needed for 350000 house is
The area covered is mathematically represented as
Answer:
There is no induced current
Explanation:
[Find the attachment]
