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3 years ago

Describe alternating current and direct current. Include two ways that they are alike and one way that they are different.

Physics

2 answers:

Katen [24]3 years ago

8 0

<h2>Answer: </h2>

<u>A direct current or DC is when the current is a constant fixed number. An alternating current or AC is when the current is alternating in a periodic manner. </u>

<h2>Explanation: </h2><h3>Similarities </h3>

Both have the flow of electrons which produce current
Both are the source of providing potential difference

<h3>Differences: </h3>

Direct current flows in one direction while alternating current flows in both directions.

ikadub [295]3 years ago

5 0

Here is your answer:

1. A alternating current is a current "is an electric current which periodically reverses direction." A direct current is a "<span>current which flows only in one direction."

2. They are alike because both are "they both are able to travel in different directions." How they are not alike is that a "alternating current travels in a reverse direction but a direct current can only travel in one direction each current."

Hope this helps!</span>

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3 years ago

A train moving at 5 m/sec passed a track gang and then accelerated uniformly a rate of 1.2 m/sec/sec for 5 min. How far did the

Amanda [17]

Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

S = 1500 m + .6 * 90000 m = 55,500 m

Check: V0 = 5 m/s

V2 = V0 + a t = 5 + 1.2 * 300 = 365 m/s

Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s (note uniform motion)

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We calculated V2 above at 365 m/s the speed after 300 sec

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3 years ago

What is the impulse

vichka [17]

Impulse is the integral of a force, F.
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3 years ago

At the same moment from the top of a building 3.0 × 10 2 m tall, one rock is dropped and one is thrown downward with an initial

Bess [88]

The equation that relates distance, velocities, acceleration, and time is,
d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and
g is the acceleration due to gravity (equal to 9.8 m/s²)

(1) Dropped rock,
(3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s

(2) Thrown rock with V₀ = 26 m/s
(3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s

The difference between the tim,
difference = 24.73 s - 5.61 s
difference = 19.12 s

<em>ANSWER: 19.12 s</em>

5 0

3 years ago

Read 2 more answers

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
$

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.

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3 years ago