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erica [24]
2 years ago
5

N the diagram, the arrow shows the movement of electric charges through a wire connected to a battery.

Physics
1 answer:
dolphi86 [110]2 years ago
5 0

Answer:

D. A difference in resistance

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If we use 1 millimeter to represent 1 light-year, how large in diameter is the Milky Way Galaxy?
storchak [24]

Answer:

d.100 meters

Explanation:

The diameter of the Milky Way Galaxy is approximately 100,000 light years.

Here we are using 1 millimiter (1 mm) to represent 1 light-year (1 ly). So, we can set the following proportion:

1 mm : 1 ly = x : 100,000 ly

and by finding x, we find the diameter of the Milky Way Galaxy in the scale used:

x=\frac{(1mm )(100,000 ly)}{1 ly}=100,000 mm = 100 m

so the correct answer is

d. 100 meters

4 0
3 years ago
A box is sliding down a ramp how many force vectors does the box have
vlada-n [284]
Well you’d have a force due to gravity, the normal force which will be perpendicular to the sources (meaning you’ll have components to this vector), and you’d have the force of friction opposing the motion of the box. I’m also assuming there’s no air resistance. In this case you’d have three vector forces.
6 0
3 years ago
Yamin is running 50 feet of No. 14 wire (with a cross section of 4,110 cmils) to a load that draws current of 11 amps. What appr
castortr0y [4]

Resistance per 1000 feet for gauge 14 wire is given as

R = 2.525 ohm

now if wire is of length 50 feet only then the resistance is given as

R = \frac{2.525}{1000}\times 50

R = 0.126 ohm

now if 11 A current flows through the wire then the voltage drop is given by ohm's law

V = iR

V = 11 \times 0.126

V = 1.4 Volts

so most appropriate answer in given options is

A. 1.8 Volts

8 0
3 years ago
[Assume g = 10m/s?] If a girl is running along a straight road with a uniform velocity 1.5 m/s, find her acceleration. (Ans: 0)
DerKrebs [107]

Answer:

Her acceleration is 0 m/s²

Explanation:

We note that the motion of the girl is on a straight road, therefore;

The vertical acceleration (e.g. due to gravity, <em>g)</em> on the horizontal motion = 0

The horizontal acceleration, a = (Change in velocity, Δv)/(Change in time, Δt)

For uniform velocity, the change in velocity, Δv = 0

Therefore, fore any change in time, Δt, we have;

a = Δv/Δt

Her acceleration, a = 0/Δt = 0

Her acceleration, a = 0 m/s²

7 0
2 years ago
A 21.3 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.7 cm from
Bezzdna [24]

Answer:

The magnetic field strength due to current flowing in the wire is9.322 x 10⁻⁶ T.

Explanation:

Given;

electric current, I = 21.3 A

distance of the magnetic field from the wire, R = 45.7 cm = 0.457 m

The strength of the resulting magnetic field at the given distance is calculated as;

B = \frac{\mu_o I}{2\pi R}

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A

B = \frac{\mu_o I}{2\pi R}\\\\B = \frac{4\pi*10^{-7} *21.3}{2\pi(0.457)}\\\\B = 9.322 *10^{-6} \ T

Therefore, the magnetic field strength due to current flowing in the wire is 9.322 x 10⁻⁶ T.

7 0
3 years ago
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