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3 years ago

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Technician A says if a fuse burns out when a switch is turned on, a short circuit is present before the switch. Technician B say

s if a fuse burns out when a switch is turned on, a short circuit is present after the switch. Who is correct?

1 answer:

lord [1]3 years ago

7 0

Answer:

Technician B

Explanation:

This can be explained as a fuse is a protective device which blows in order to protect the circuit and the equipment from short circuiting.

Short circuit is a condition where a very heavy current flow through the line with no low line impedance which is not the intended path for the current to flow.

Thus when heavy current flows through a fuse it blows out thus disconnecting the path for the current flow.

Therefore, if a fuse burns out when switch is on, short circuit is present after the switch

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So, what is the answer of this assignment?

Ket [755]

What assignment, you have to work with me here i don't know what were working on

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3 years ago

What is the charge on 1.0 kg of protons? (e = 1.60 × 10-19 c, mproton = 1.67 × 10-27 kg)?

Andru [333]

First, we need to find the number of protons, which is the total mass divided by the mass of one proton:
$N= \frac{m}{m_p}= \frac{1.0 kg}{1.67 \cdot 10^{-27} kg}=6.0 \cdot 10^{26}$ protons

Then, the total charge is the number of protons times the charge of a single proton:
$Q=Ne = 6.0 \cdot 10^{26}\cdot 1.60 \cdot 10^{-19} C=9.6 \cdot 10^7 C$

8 0

3 years ago

Read 2 more answers

A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?

posledela

Answer:

$118\; \rm Hz$ .

Explanation:

The frequency $f$ of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of $v= 295\; \rm m\cdot s^{-1}$ . In other words, the wave would have traveled $295\; \rm m$ in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that $295\; \rm m$ ? The wavelength of this wave $\lambda = 2.50\; \rm m$ gives the length of one wave cycle. Therefore:

$\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118$ .

That is: there are $118$ wave cycles in $295\; \rm m$ of this wave.

On the other hand, Because that $295\; \rm m$ of this wave goes through that point in each second, that $118$ wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

$f = 118\; \rm s^{-1} = 118\; \rm Hz$ .

The calculations above can be expressed with the formula:

$\displaystyle f = \frac{v}{\lambda}$ ,

where

$v$ represents the speed of this wave, and
$\lambda$ represents the wavelength of this wave.

6 0

3 years ago

Please help if you can.

VMariaS [17]

Initially there were 10 bulbs of 60 Watt power

So total power of all bulbs = 60 * 10 = 600 W

now each bulb used for 4 hours daily

so total energy consumed daily

$E = P * t$

$E = 600 * 4 = 2400 Wh$

$E = 2.4 kWh$

now we have total power consumed in 1 year

$E = 365 * 2.4 = 876 kWh$

cost of electricity = 10 cents/ kWh

so total cost of energy for one year

$P_1 = 876 * 10 = 8760 cents = $87.60$

Now if all 60 Watt bulbs are replaced by 30 Watt bulbs

So total power of all bulbs = 30 * 10 = 300 W

now each bulb used for 4 hours daily

so total energy consumed daily

$E = P * t$

$E = 300 * 4 = 1200 Wh$

$E = 1.2 kWh$

now we have total power consumed in 1 year

$E = 365 * 1.2 = 438 kWh$

cost of electricity = 10 cents/ kWh

so total cost of energy for one year

$P_1 = 438 * 10 = 4380 cents = $43.80$

total money saved in 1 year

$Savings = 87.6 - 43.8 = $43.80$

3 0

3 years ago

A person kicks a ball off of a 50m high cliff with a speed of 10 m/s. How long will it take the ball to hit the ground? * 7 poin

Musya8 [376]

Presumably, the ball is kicked parallel to the ground below the cliff, so its altitude <em>y</em> at time <em>t</em> is

$y(t)=50\,\mathrm m-\dfrac12gt^2$

where <em>g</em> = 9.80 m/s^2 is the acceleration due to gravity.

The ball hits the ground when <em>y</em> = 0:

$0 = 50\,\mathrm m-\dfrac12gt^2$

$t^2=\dfrac{100\,\mathrm m}g$

$t=\dfrac{10}{9.80}\,\mathrm s\approx\boxed{3.2\,\mathrm s}$

6 0

3 years ago