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Illusion [34]
1 year ago
15

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

e loose and touch the metal case.B If the metal case becomes live, the earth wire conducts current to the ground.C The earth wire needs to have a high resistance.D Earthing metal cases helps prevent a person from receiving an electric shock.
Physics
1 answer:
labwork [276]1 year ago
7 0

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.

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Radda [10]

Explanation:

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4 0
3 years ago
Two slits are illuminated with green light (λ = 540 nm). The slits are 0.05 mm apart and the distance to the screen is 1.5 m. At
zhenek [66]

Answer:

0.21486 mm

Explanation:

The formula for the maximum intensity is given by;

I = I_o•cos²(Φ/2)

Now,we are not given Φ but it can be expressed in terms of what we are given as; Φ = πdy/(λL)

Where;

y is the distance from the central maximum

d is the distance between the slits

λ is the wavelength

L is the distance to the screen

Thus;

I = I_o•πdy/(λL)

We are given;

d = 0.05 mm = 0.5 × 10^(-3) m

λ = 540 nm = 540 × 10^(-9) m

L = 1.25 m

I/I_o = 50% = 0.5

From earlier, we saw that;

I = I_o•πdy/(λL)

We have I/I_o = 0.5

Thus;

I/I_o = πdy/(λL)

Plugging in the relevant values;

0.5 = (π × 0.5 × 10^(-3) × y)/(540 × 10^(-9) × 1.25)

Making y the subject, we have;

y = (0.5 × 540 × 10^(-9) × 1.25)/(π × 0.5 × 10^(-3))

y = 0.00021486 m

Converting to mm, we have;

y = 0.21486 mm

7 0
3 years ago
Water drips from the nozzle of a shower onto the floor 190 cm below. The drops fall at regular (equal) intervals of time, the fi
VARVARA [1.3K]

Answer:

Second drop: 1.04 m

First drop: 1.66 m

Explanation:

Assuming the droplets are not affected by aerodynamic drag.

They are in free fall, affected only by gravity.

I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.

We can use the equation for position under constant acceleration.

X(t) = x0 + v0 * t + 1/2 * a *t^2

x0 = 0

a = 9.81 m/s^2

v0 = 0

Then:

X(t) = 4.9 * t^2

The drop will hit the floor when X(t) = 1.9

1.9 = 4.9 * t^2

t^2 = 1.9 / 4.9

t = \sqrt{0.388} = 0.62 s

That is the moment when the 4th drop begins falling.

Assuming they fall at constant interval,

Δt = 0.62 / 3 = 0.2 s (approximately)

The second drop will be at:

X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m

And the third at:

X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m

The positions are:

1.9 - 0.86 = 1.04 m

1.9 - 0.24 = 1.66 m

above the floor

8 0
3 years ago
A. A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. What is the maximum height h that it achieves, and w
sveta [45]

Answer:

A)     t = 0.40816 s , y = 0.916 m

Explanation:

A) For this problem we use the kinematic relations

           v = v₀ - g t

the highest point zero velocities (v = 0)

           t = (v₀-v) / g

           t = (4 - 0) / 9.8

           t = 0.40816 s

to calculate the height let's use

          v² = v₀² - 2 g y

          y = vo2 / 2g

           y = 4 2 / (2 9.8)

          y = 0.916 m

To find the number of photos, we can use a direct proportions rule, if you take 30 photos in a second in 0.40816 s how many photos does it take

          # _photos1 = 0.40916 (30/1)

          # _photos1 = 12

yes i take 120 fps

          #_fotod = 0.40916 (120/1)

          #photos = 5.87 10³

 

B) The ball is released from a latura h how long it takes to reach the floor

           v² = v₀² + 2 g y

where the initial velocity is zero and the velocity with which the expert leaves is equal to the velocity with which v = 4 m / s leaves

            v² = 2gy

             v = √ (2 9.8 0.916)

             v = √ (2.1397 101)

             v = 4.6257 m / s

c) we ask us for the time for latura

             y = L / 2

             y = 0.916 / 2

             y = 0.458 m

now we can use the formula

             y = v₀ t - ½ g t²

           0.458 = 4.00 t - ½ 9.8 t²

            4.9 t² - 4t + 0.458 = 0

            t² -0.8163 t +0.09346 = 0

we solve second degree execution

           t = [0.8163 ±√ (0.8163² - 4 0.09346)] / 2

           t = [0.8163 ± 0.540] / 2

           t₁ = 0.678 m

           t₂ = 0.2763 m

the shortest time is for when the ball goes up and the longest when it goes down

D) the graph of vs Vs is expected to be a closed line

and the graph of position versus time a parabola

0 0
4 years ago
when an element tends to lose its valence electrons in chemical reactions , does it behave more like a metal or nonmetal
Juli2301 [7.4K]

It behaves more like a metal

Explanation:

When an element tends to lose its valence electrons in chemical reactions, they behave more like a metal.

Metals are electropositive.

Electropositivity or metallicity is the a measure of the tendency of atoms of an element to lose electrons.

This is closely related to ionization energy and the electronegativity of the element.

  • The lower the ionization energy of an element, the more electropositive or metallic the element is .

Metals are usually large size and prefers to be in reactions where they can easily lose their valence electrons.

When most metals lose their valence electrons, they attain stability.

Non-metals are electronegative. They prefer to gain electrons.

learn more:

Reactivity brainly.com/question/6496202

#learnwithBrainly

4 0
4 years ago
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