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3 years ago

Why are the outlets in homes never wired in series? What problems might this present?

Physics

2 answers:

Yuri [45]3 years ago

7 0

Answer:

It would be more uncomfortable to use electricity in your house as all the outlets must be connected to appliances so the electricity can flow.

Explanation:

Series circuits in real-life

Time ago christmas lights were wired in series, it means electricity flowed in one single way through all the bulbs. The problem was when one of those many bulbs stopped working because it interrupmted the only path for eletricity and all the bulbs turned off as only one of them is broken. Very annoying, isn't it? Now lets repeat the example imagining the outlets of your livingroom connected in series. Let say you have 4 outlets connected in series. The main problem is that each one of those outlets works as a broken bulb as long as no appliance is connected because electricity must flow from the first outlet to appliance and then it will go to the second outlet and so on until reaching the 4th outlet and the 4th appliance. It works that way because there is only one way for electriciy to flow through the 4 outlets. pretty unpractical, isn't it? let see a better solution

Parallel outlets

The good thing about parallel outlets is that they are directly connected to the power supply and the only element that is necessary to let the electricity fllow is the appliance. Each outlet is part of a different way (a different circuit) so you can manage your outlets separately. Attached you will find a picture to ilustrate this better. Remember that each outlet in series works exactly like a broken bulb. If you have any doubts, just let me know.

rosijanka [135]3 years ago

3 0

A series circuit means that there is only one current. Which means that everything is sharing that one wire. Homes are normally wired with a parallel circuit, which means that there is more than one current running through the house. If you have a series, if you turn on one thing for example a light switch, than everything will turn on with it. But of you have a parallel you can turn on everything one by one, and save energy.

Trust me. I just took my semester test in physical science this morning, and I passed. This is exactly what we learned.

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A rightward force is applied to a box in order to move it across the table at a constant velocity. (Ignore wind resistance).

anzhelika [568]

The forces would be gravity and friction

3 0

3 years ago

For a vehicle to negotiate a banked curve in poor weather conditions, where the force of friction f = 0, for a given velocity v

djverab [1.8K]

Answer:

$R=240m$

Explanation:

From the question we are told that:

Velocity $v=25m/s$

Force of friction f = 0

Angle $\theta=15$

Generally the equation for Radius of curvature is mathematically given by

$R=frac{v^2}{tan\theta *g}$

$R=frac{25^2}{tan 15 *9.81}$

$R=240m$

7 0

3 years ago

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?

timurjin [86]

Answer: The drag force goes up by a factor of 4

Explanation:

The Drag Force equation is:

$F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2}$ (1)

Where:

$F_{D}$ is the Drag Force

$C_{D}$ is the Drag coefficient, which depends on the material

$\rho$ is the density of the fluid where the bicycle is moving (air in this case)

$A_{D}$ is the transversal area of the body or object

$V$ the bicycle's velocity

Now, if we assume $C_{D}$ , $\rho$ and $A_{D}$ do not change, we can rewrite (1) as:

$F_{D}=C.V^{2}$ (2)

Where $C$ groups all these coefficients.

So, if we have a new velocity $V_{n}$ , which is the double of the former velocity:

$V_{n}=2V$ (3)

Equation (2) is written as:

$F_{D}=C.V_{n}^{2}=C.(2V)^{2}$

$F_{D}=4CV^{2}$ (4)

Comparing (2) and (4) we can conclude the Drag force is four times greater when the speed is doubled.

7 0

3 years ago

Please help with these questions as well! I need urgent help! I will give brainliest! God bless!

Radda [10]

6. Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object. So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.

7. Same idea as question 2.

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N

So the person is exerting 144 N.

10. First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left. However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A. The force that is causing block B to experience the lower relative force to the right is because of the friction. To find the friction:

The sum of the forces in the leftward and rightward direction for block B must equal 12N.

75 - x = 12

x = 63N

So the force of friction of block A on block B is 63N to the left.

5 0

3 years ago

Potential difference of a battery is 2.2 V when it is connected

Alchen [17]

Answer:

1.1ohms

Explanation:

According to ohms law E = IR

If potential difference of a battery is 2.2 V when it is connected across a resistance of 5 ohm and if suddenly the voltage Falls to 1.8V then the current in the 5ohms resistor I = V/R = 1.8/5

I = 0.36A (This will be the load current).

Before we can calculate the value of the internal resistance, we need to know the voltage drop across the internal resistance.

Voltage drop = 2.2V - 1.8V = 0.4V

Then we calculate the internal resistance using ohms law.

According to the law, V = Ir

V= voltage drop

I is the load current

r = internal resistance

0.4 = 0.36r

r = 0.4/0.36

r = 1.1 ohms

6 0

3 years ago