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4 years ago

Whats the difference between watts, amps, and volts?

Physics

1 answer:

daser333 [38]4 years ago

6 0

Answer:

Watts is Power (W)

Amps is current (A)

Volts is Voltage (v)

Explanation:

Watts is a derived unit of 1 joule per second, and is used to quantify the rate of energy transfer. In SI (standard international) base units, the watt is described as kg⋅m²⋅s⁻³, which can be demonstrated to be coherent by dimensional analysis

An ampere is the unit used to measure electric current. Current is a count of the number of electrons flowing through a circuit. One amp is the amount of current produced by a force of one volt acting through the resistance of one ohm

Volt is the electrical unit of voltage or potential difference (symbol: V). One Volt is defined as energy consumption of one joule per electric charge of one coulomb. 1V = 1J/C. One volt is equal to current of 1 amp times resistance of 1 ohm: 1V = 1A ⋅ 1Ω

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-List ten different waves that you encounter in everyday life. Be specific, don't just say "sound

Nady [450]

Answer:

1.Radio waves from using your TV.

2. microwaves you are in satellites.

3.UV light waves come from sunlight

4.x ray waves used at the doctors office

5. infrared waves are in remote controls when sending signals.

I don't know if I can think of another 5 right now. but I hope this helps.

6 0

3 years ago

Once ignited, a small rocket motor on a spacecraft exerts a constant force of 10 N for 7.80s. During the burn the rocket causes

Marat540 [252]

Answer:

a = 0.01 [m/s²]

Explanation:

To solve this problem we must use Newton's second law, which says that the sum of forces on a body is equal to the product of mass by acceleration.

ΣF = m*a

where:

F = forces = 10 [N]

m = mass = 100 [kg]

a = acceleration [m/s²]

10 = 100*a

a = 0.01 [m/s²]

4 0

4 years ago

What use do we have for motion diagrams?

mamaluj [8]

Answer:

A motion diagram represents the motion of an object by displaying its location at various equally spaced times on the same diagram. Motion diagrams are a pictorial description of an object's motion. They show an object's position and velocity initially and present several spots in the center of the diagram.

Explanation:

7 0

3 years ago

What is the name of the plate to the west and directly adjacent to the plate on which Chile is located?

Ksenya-84 [330]

<h2>Answer: Nazca plate</h2>

Explanation:

The Nazca plate is an oceanic tectonic plate that is found in the Pacific Ocean off the west coast of South America, specifically in front of the north and central coast of Chile and the entire coastline of Peru, Ecuador and Colombia. This plate is in constant motion, which causes it to sink under the South American (phenomenon known as subduction).

It is mainly because of this subduction process that this region has a lot of seismic activity. Another important aspect is that thanks to these movements, the Andes mountain range and the Peruvian-Chilean fossa originated.

6 0

3 years ago

A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a

Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

$Z_{R}$ = R

R = resistance

$Z_{L}$ = jωL

ω = voltage source angular frequency, L = inductance

$Z_{C}$ = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

$Z_{R}$ = 217Ω

$Z_{L}$ = j(220)(0.875) = j192.5Ω

$Z_{C}$ = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$ + $Z_{C}$

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

$v_{R}$ = I× $Z_{R}$

$v_{L}$ = I× $Z_{L}$

$v_{C}$ = I× $Z_{C}$

Given values:

I = (0.0569∠1.147+220t)A

$Z_{R}$ = 217Ω = (217∠0)Ω

$Z_{L}$ = j192.5Ω = (192.5∠π/2)Ω

$Z_{C}$ = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

$v_{R}$ = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

$v_{L}$ = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

$v_{C}$ = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

$v_{R}$ = (12.35∠1.147+220t)V

$v_{L}$ = (10.95∠2.718+220t)V

$v_{C}$ = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

$v_{R}$ = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

$v_{L}$ = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

$v_{C}$ = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0

3 years ago