<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>
Weight of the child m = 50 kg
Radius of the merry -go-around r = 1.50 m
Angular speed w = 3.00 rad/s
(a)Child's centripetal acceleration will be a = w^2 x r = 3^2 x 1.50 => a = 9 x
1.5
Centripetal Acceleration a = 13.5m/sec^2
(b)The minimum force between her feet and the floor in circular path
Circular Path length C = 2 x 3.14 x 1.50 => c = 3 x 3.14 => C = 9.424
Time taken t = 2 x 3.14 / w => t = 6.28 / 3 => t = 2.09
Calculating velocity v = distance / time = 9.424 / 2.09 m/s => v = 4.5 m/s
Calculating force, from equation F x r = mv^2 => F = mv^2 / r => 50 x (4.5)^2
/ 1.5
F = 50 x 3 x 4.5 => F = 150 x 4.5 => F = 675 N
(c)Minimum coefficient of static friction u
F = u x m x g => u = F / m x g => u = 675/ 50 x 9.81 => 1.376
u = 1.376
Hence with the force and the friction coefficient she is likely to stay on merry-go-around.
Answer:
velocity during second d = 20.0 mi/h
Explanation:
Total distance travelled is 2d, with an average velocity of 30.0 mi/h you can express the time travelled in terms of d:
distance = velocity * time
time = distance / velocity
time = 2d/30.0
The time needed for the first d at 60.0 is:
time = d/60.0
The time in the second d you can get it by substracting both times (total time - time for the first d)
second d time = 2d/30.0 - d/60.0
= 4d/60.0 - d/60.0
= 3d/60.0
and with the time (3d/60.0) and the distance travelled (d) you can get the velocity:
velocity = distance / time
velocity = d / (3d/60.0)
= 60.0/3 = 20.0 mi/h
Answer:
Elliptical galaxies
Explanation:
Edwin Hubble classified galaxies into three categories
Elliptical
Spiral
Lenticular
The elliptical galaxies have an elipsoidal shape roughly. They have stars which are old and the primary light source of the galaxy. The formation of new stars is very limited. This increases the brightness of the galaxy. The mass of the stars are low. So, far the percentage elliptical galaxies is low compared to other galaxies.
