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3 years ago

How are parallel circuits used in homes?

1 answer:

5 0

Every single thing in your home that's plugged into a wall outlet or works with a switch on the wall . . . every lamp, ceiling light, room fan, computer, microwave, stove, hair dryer, coffee pot, toaster, slow cooker, TV, home entertainment system, printer, wifi router, cordless phone, phone charger, and clock radio ... they're ALL in parallel.

<u>And</u> chances are that everything in your house is also in parallel with everything in the houses of a few of your neighbors on each side of you.

THAT's how parallel circuits are used in residential areas.

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A student completed a lab report. Which correctly describes the difference between the "Question" and "Hypothesis" sections

galben [10]

Answer:

the second one!

Explanation:

the question is well, the question, a hypothesis is an educated guess on what you think will be the outcome

4 0

3 years ago

Read 2 more answers

If at 10m above the ground an object has 50J of Kinetic Energy, with 50J of Potential Energy. How high can the object travel?

steposvetlana [31]

Hi there!

$\large\boxed{\text{B) 20 meters}}$

We know that:

$E_T = U + K$

U = Potential Energy (J)

K = Kinetic Energy (J)

E = Total Energy (J)

At 10m, the total amount of energy is equivalent to:

U + K = 50 + 50 = 100 J

To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:

100J = mgh

We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².

50J = 10(10)m

m = 1/2 kg

Now, solve for height given that E = 100 J:

100J = 1/2(10)h

100J = 5h

<u>h = 20 meters</u>

3 0

2 years ago

If a car has a kinetic energy of 40000 J and is moving at a velocity of 25 m/s, what is the mass of the car? KE=1/2mv

ohaa [14]

Answer:

3200 Kg

Explanation:

Just apply the formula:

40.000 = 1/2 . m . 25

80.000 = 25m

m = 80.000/25

m = 3200 Kg

7 0

3 years ago

A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?

anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}$

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

$\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}$

now, replace

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}$

hence, the answer is 0.495 m/s

3 0

1 year ago

Using the same cost and time estimates, consider the time-effectiveness of each engineer.

sammy [17]

Answer:

Camilla

Explanation:

I got it right on edge. :)

4 0

3 years ago