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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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> Which activity will improve your muscle strength

BartSMP [9]

Answer:

Lifting weights
Strength training
Resistance training
Heavy gardening e.g digging , shovelling

Explanation:

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3 years ago

Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations wa

OLga [1]

Answer:

Explanation:

The equation for a traveling wave as expressed as y(x, t) = A cos(kx − $\omega$ t) where An is the amplitude f oscillation, $\omega$ is the angular velocity and x is the horizontal displacement and y is the vertical displacement.

From the formula; $k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f$ where;

$\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency$

Before we can get the transverse speed, we need to get the frequency and the wavelength.

frequency = 1/period

Given period = 2/15 s

Frequency = $\frac{1}{(2/15)}$

frequency = 1 * 15/2

frequency f = 15/2 Hertz

Given wavelength $\lambda$ = 2m

Transverse speed $v = f \lambda$

$v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s$

Hence, the transverse speed at that point is 15m/s

8 0

3 years ago

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

r < a:

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

a < r < b:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

r = b:

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

b < r < c:

E = 0

r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

r < c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

r<a:

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

At r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

At a < r < b:

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

$E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = b:

$E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

Again, the minus sign indicates the direction of the field towards the center.

At b < r < c:

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

At r < c:

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago

according to newton's third law, when a horse pulls on a cart, the cart pulls back on the horse with an equal force on the horse

Debora [2.8K]

Based on Newton's principle, whenever objects A and B interact with each other, they exert forces upon each other.

When a horse pulls on a cart, the horse exerts a force only to the cart. But that force applies only to the cart, not to the horse.

The cart in turn exerts a force on the horse. But that force applies only to the horse, not the cart also.

There are two forces resulting from this interaction - a force on the horse and a force on the cart. The net force on the cart remains as it was --- a positive force in the direction of the horse's movement. Therefore, the cart begins to accelerate and move.

8 0

3 years ago

A baseball is thrown with an initial velocity of 45.4 m/s at an angle of 31.2 ∘ .

sveticcg [70]

Answer:

V (initial vertical velocity) = 45.4 sin 31.2 = 23.52 m/s

1/2 m V^2 = m g h conservation of energy

h = V^2 / (2 g) = 23.52^2 / 19.6 = 28.2 m max height

Check:

t = 28.2 / 9.8 = 2.88 sec time to reach max height

h = 23.52 * 2.88 - 1/2 g 2.88^2 = 27.1 m

8 0

2 years ago