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The metal cases of electrical appliances are connected to an earth wire.Which statement is not correct?A The live wire may becom

labwork [276]

Given that metal cases of electrical appliances are connected to the wire, let's select the statement which is not correct from the list of statements.

The earth wire is used to protect you and help reduce the risk of receiving an electric shock. The earth wire reduces the risk of electric shock by creating a path for a fault or lose current to flow to the Earth.

The Live wire may become loose and touch the metal case. In this case, the earth wire will channel the fault current to the earth thereby reducing the risk of electric shock.

If the metal case becomes live, the earth wire conducts current to the ground. This helps prevent electric shock from the metal case.

The Earth wire has low or no resistance. It is always made of copper.

It provides a low resistance path to the ground.

Therefore, the statement ''the earth wire needs to have high resistance'' is NOT current.

ANSWER:

C. The earth wire needs to have a high reistance.

7 0

1 year ago

When hydrogen is fused into helium, energy is released from Choose one: A. the increase in pressure. B. the decrease in the grav

Tatiana [17]

The nuclear fusion of hydrogen atoms releases a huge amount of energy. So the correct choice is C. Conversion of mass to energy.

What is nuclear fusion?

When two small nuclei join to form a new nucleus, then this process is termed nuclear fusion. A huge amount of energy is released when there occurs nuclear fusion between the two nuclei. And a new element is formed.

It has been observed that the amount of energy released in nuclear fusion is equal to the mass difference between the mass of the formed nucleus and the total mass of old nuclei. Hence in the nuclear fusion of hydrogen nuclei to form a helium nucleus, the energy is released due to the conversion of mass into energy.

The pressure is increased to make the hydrogen atoms fuse but this change in pressure does not contribute to the energy released in the fusion of hydrogen.

The magnitude of the gravitational field is too low and it does not contribute to the energy released in the fusion of hydrogen.

The gravitational collapse does not occur between the two hydrogen atoms. This phenomenon occurs in celestial bodies so this also does not contribute to the energy released in the fusion of hydrogen.

Learn more about nuclear fusion here:

brainly.com/question/10165218

#SPJ4

8 0

1 year ago

How should you behave while performing laboratory experiments?

Brrunno [24]

Explanation:

Always behave responsibly in the laboratory. Do not run around or play practical jokes. Always check the safety data of any chemicals you are going to use. Never smell, taste or touch chemicals unless instructed to do so.

4 0

3 years ago

Read 2 more answers

Why is a protective apron or lab coat important to use when working with acids?

Kipish [7]

Answer:

Acids break down fabrics and can cause burns if the acids are strong.

Explanation:

A protective apron or lab coat is important when working with acids because acids break down fabrics and can cause burns if the acids are strong.

An acid is a substance that interacts with water to produce excess hydroxonium ions in an aqueous solution.
A strong acid ionizes completely in solution.
When they come in contact with a fabric, they break them down violently.
So, if they come in contact with the skin, it causes a violent break down of body tissues.
The apron acts a protective layer.

4 0

3 years ago

Read 2 more answers

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