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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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3 years ago
Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1306p(1/2), where p is the pressure (in psi). If th
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Answer:

a=38.5 ft/sec^{2}

Explanation:

Note that acceleration is the rate change of velocity i.e

acceleration=\frac{change in velocity}{change in time}\\a=\frac{dv}{dt} \\.

Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

\frac{dv}{dt}=1306*(1/2)p^{-1/2}\frac{dp}{dt} \\

if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have

a=653*0.1667*0.354\\

a=\frac{dv}{dt}=653(36)^{-1/2}*0.354\\  a=38.5 ft/sec^{2}

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3 years ago
Can any juniors who go to Texas connections academy help me out with physics?
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3 years ago
A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in
german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a (Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

m - Mass of the box, measured in kilograms.

a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_{o}+a\cdot t), we expand the previous expression:

-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right) (Eq. 1b)

Where:

W - Weight, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,\frac{m}{s^{2}}, v_{o} = 1.37\,\frac{m}{s}, v = 0\,\frac{m}{s} and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s}  }{2.8\,s} \right)

f = 2.614\,N

The magnitude of the friction force exerted on the box is 2.614 newtons.

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3 years ago
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