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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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A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go? (b) how long is it in the air

Mariulka [41]

Refer to the diagram shown below.

When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m

Answer: The maximum height attained is 24.7 m (nearest tenth)

Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s

To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.

Answer: The ball stays in the air for 4.5 s (nearest tenth)

3 0

4 years ago

a football player initially at rest accelerates uniformly as she runs down the field, travelling 17m (E) in 3.8s. What is her fi

ikadub [295]

X - Xo = 0.5(Vo +V)t

Xo cancels because it is 0 (she is at rest)

17 = 0.5(0+V)3.8

17 = 1.9V

Final velocity is 8.94737 m/s

6 0

3 years ago

Describe the image of an object that is 38 cm from a concave mirror that has a focal length of 10 cm.

harina [27]

Let's start by using the mirror equation:
$\frac{1}{f}= \frac{1}{d}+ \frac{1}{q}$
where f=10 cm is the focal lenght of the mirror, d=38 cm is the distance of the object from the mirror, while q is the distance of the image from the mirror.
For the sign convention, f is taken as positive for a concave mirror. Therefore, we can solve the equation for q:
$\frac{1}{q}= \frac{1}{f}- \frac{1}{d} = \frac{1}{10 cm}- \frac{1}{38 cm}$
from which we find q=13.6 cm.

The fact that q is positive means that the image is real, so it is on the same side of the object, with respect to the mirror.

Then we can also find the size of the image with respect to the original object. The magnification is given by
$M=- \frac{q}{d} =- \frac{13.6}{38}=-0.36$
The negative sign means that the image is inverted, and the size of the image is 0.36 times the size of the object.

4 0

3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

3 years ago

1. Write the following degrees in radian measure

svp [43]

C is the right answer

I hooe this helps

6 0

3 years ago