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3 years ago

This diagram shows a process the power stars. This process is called

Physics

2 answers:

goldenfox [79]3 years ago

8 0

its c........................................................................

Damm [24]3 years ago

6 0

I think its oxidation.

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A major difference between AC and DC electricity is that A) the current in AC electricity varies in magnitude and direction. B)

galina1969 [7]

Answer:

A) the current in AC electricity varies in magnitude and direction.

C) the voltage in AC electricity varies in magnitude and direction.

Explanation:

In DC current and voltage the direction of current will not change with time and it always remains the same.

So here in DC voltage and DC current the magnitude may change with time but the direction will always remain same

While in AC voltage and AC current the direction of AC will change with time

periodically.

So here magnitude and direction both will change in AC current and AC voltage.

so the correct answer is

A) the current in AC electricity varies in magnitude and direction.

C) the voltage in AC electricity varies in magnitude and direction.

7 0

3 years ago

Read 2 more answers

What is the difference between a neap and a spring tide in terms of size?

Mamont248 [21]

Spring tides occur when the moon is either new or full, and the sun, the moon, and the Earth are aligned. ... neap tide- A tide in which the difference between high and low tide is the least. Neap tides occur twice a month when the sun and moon are at right angles to the Earth.

7 0

3 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago

A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

4 0

3 years ago

The charge of an electron is

AlladinOne [14]

Proton positive; electron negative; neutron no charge. The charge on the proton and electron are exactly the same size but opposite. The same number of protons and electrons exactly cancel one another in a neutral atom.

hoped it helped

6 0

3 years ago