What happens to the intensity of solar energy as latitude increases

A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire?

andrey2020 [161]

Answer: The correct option is A ( horizontally towards the east)

Explanation:

Magnetic field is a region around a magnet or a current- carrying conductor, where a magnetic force is experienced. The magnetic effect of electric current was first discovered in the early 1820 by Oersted. Using a wire that had current flowing through it and a pivoted magnetic needle, he discovered that the direction of deflection depended on the direction of the current and whether the wire was above or below the needle.

From the way the needle turns when current when current carrying wire is held parallel to it, he therefore concluded that:

--> a current has magnetic field all round it,

--> the magnetic field is in a direction perpendicular to the current.

The above discovery was now modified in Fleming's left hand rule which states that when conductor carrying current is placed in a magnetic field, the conductor will experience a force perpendicular to both the field and the flow of current.

Therefore from the question, a vertical wire carrying current in DOWNWARD direction is placed in a HORIZONTAL magnetic field directed to the NORTH. The direction of the force on the wire is to the EAST.

4 0

3 years ago

A 1.11 kg piece of aluminum at 78.3 c is put into a glass with 0.210 kg of water at 15.0

ValentinkaMS [17]

M = mass of aluminium = 1.11 kg

$c_{a}$ = specific heat of aluminium = 900

$T_{ai}$ = initial temperature of aluminium = 78.3 c

m = mass of water = 0.210 kg

$c_{w}$ = specific heat of water = 4186

$T_{wi}$ = initial temperature of water = 15 c

T = final equilibrium temperature = ?

using conservation of heat

Heat lost by aluminium = heat gained by water

M $c_{a}$ ( $T_{ai}$ - T) = m $c_{w}$ (T - $T_{wi}$ )

(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)

T = 48.7 c

7 0

3 years ago

2. An electron and a proton are separated by 5 cm:

Diano4ka-milaya [45]

Answer:

a) the charge of an electron is equivalent to the magnitude of the elementary charge but barring a negative sign since the side of the elementary charge is roughly 1.602 * 10 - 19 Columbus then the charge of the electronic is-1.602 * 10 - 19

b) b=2T on the electron moving in the magnetic field

7 0

3 years ago

The windshield of a car has a total length of arm and blade of 9 inches, and rotates back and forth through an angle of 93degre

Sergio039 [100]

Answer:

The area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

Explanation:

Given

Length of blade = 9 inches

Angle of rotation = 93°

We're to calculate the area of the portion of the windshield cleaned by the 7-in wiper blade?

We'll solve this by using area of a sector.

Area of a sector = ½r²θ

where θ is in radians.

So, angle of rotation (93°) must first be converted to radians

Converting 93º to radians, we get 31π/60

The area of the region swept out by the wiper blade = (area of the sector where r = 9 and

θ = 31π/60) - (area of the sector where r = (9-7) and θ = 31π/60).

We're making use of 9-7 because that region is outside the boundary of the 7in blade

So Area = ½*9²*31π/60 - ½*2²*31π/60

Area = ½*31π/60(9²-7²)

Area = 31π/120 * (81 - 49)

Area = 31π/120 * 32

Area = 992π/120

Area = 62.49151386765697 in²

Area ≈ 62.49 in²

Hence, the area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

5 0

4 years ago

Read 2 more answers

A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th

ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a) Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0

3 years ago