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3 years ago

5

An electric space heater draws 16.0 amps from a 120 volt source. This device is operated, on average, for 8.0 hours a day. How m

uch power does the heater consume?

1 answer:

iogann1982 [59]3 years ago

6 0

Answer:

1920watts

Explanation:

P=IV

P = 16 x 120 = 1920 watts

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The PV system is operating in a location where the annual average daily incident solar energy (the insolation) on the array equa

vlabodo [156]

The average amount of solar energy incident on the PV per day is 10000 kWh/day.

<h3>Equation</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Let the PV array has an area equal to 50 square meters. Hence:

Average amount of solar energy incident on the PV per day = 200 kWh/m²/day * 50 m² = 10000 kWh/day.

Find out more on Equation at: brainly.com/question/2972832

5 0

2 years ago

how much force is needed to move a brick with a mass of 345-kg a distance of 28m while doing 1008 j of work?

Travka [436]

    Work = (force) x (distance)

   1,008 J = (force) x (28 m)

Divide each side by 28m : (1,008 kg-m²/sec²) / (28 m) = force

   Force = 36 kg-m/s² = 36 Newtons .
      (about 8.1 pounds)

It doesn't matter what that force accomplishes.
It could be moving a brick, lifting a fish, or pushing a little red wagon.
In order to do 1,008 joules of work in 28 meters, it takes 36 N of force,
in the direction of the 28 meters.

5 0

2 years ago

What is the magnetic flux density (B-field) at a distance of 0.36 m from a long, straight wire carrying a current of 3.8 A in ai

olga nikolaevna [1]

Answer:

The magnetic flux density is $2.11\times10^{-6}\ T$

Explanation:

Given that,

Distance = 0.36 m

Current = 3.8 A

We need to calculate the magnetic flux density

Using formula of magnetic field

$B =\dfrac{\mu_{0}I}{2r}$

Where,

r = radius

I = current

Put the value into the formula

$B =\dfrac{4\pi\times10^{-7}\times3.8}{2\times\pi\times0.36}$

$B=2.11\times10^{-6}\ T$

Hence, The magnetic flux density is $2.11\times10^{-6}\ T$

3 0

2 years ago

A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

8 0

3 years ago

Pls help i’ll give brainliest if you give a correct answer!!

enot [183]

Answer:

the second one

Explanation:

4 0

2 years ago