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3 years ago

4. Um chuveiro elétrico instalado em uma rede de

Physics

1 answer:

gulaghasi [49]3 years ago

6 0

Answer:

1.1 kilowatt-hour

Explanation:

The question is:

4. An electric shower installed in a network of 220 V, when in operation, is traversed by an electric current of intensity 20 A. Determine the electrical energy consumed, in kilowatt-hour, during a bath lasting 15 min.

1. Data:

a) V = 220V

b) I = 20 A

c) E = ?

2. Physical principles and formulae

Power, P, is equal to the product of the potential difference, V, by the current, I:

$P=V\times I$

The energy, E, is equal to the product of the power, P, and the time, t:

$E=P\times t$

3. Solution

$P=220V\times 20A=4,400W=4,400J/s$

$E=4,400W\times 15min\times 1hour/60min=1,100W-h=1.1kwatt-hour$

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A weather station records the wind as blowing from the northeast at 12 km/h.

weqwewe [10]

Answer:

Explanation:

vector quantities always have spedd and direction

7 0

3 years ago

On another planet, the isotopes of titanium have the given natural abundances. Isotope Abundance Mass (u) 46Ti 77.600% 45.95263

allsm [11]

Answer: Average atomic mass of titanium on that planet is 46.52

Explanation:

Mass of isotope Ti-46 = 45.95263

% abundance of isotope Ti-46 = 77.600 % = $\frac{77.600}{100}=0.776$

Mass of isotope Ti- 48= 47.94795

% abundance of isotope Ti-48 = 16.100%= $\frac{16.100}{100}=0.161$

Mass of isotope Ti- 50 = 49.94479

% abundance of isotope Ti-50 = 6.300%= $\frac{6.300}{100}=0.063$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$Z=\sum[(45.95263 \times 0.776)+(47.94795 \times 0.161)+(49.94479\times 0.063)$

$Z=46.52$

Average atomic mass of titanium on that planet is 46.52

8 0

3 years ago

Hi please, I Have An attachment on Waves, Just two Objective Questions Whoever Answers Will be Marked Brainliest thank you.

kifflom [539]

Answer:

The first answer is W and Z, since they appear to be a period apart. Dont know the second question. I did what I could, hope someone can answer the second.

5 0

3 years ago

During a tennis match, a player serves the ball at 27.4 m/s, with the center of the ball leaving the racquet horizontally 2.34 m

Maurinko [17]

a. The ball's horizontal and vertical positions at time $t$ are given by

$x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t$

$y=2.34\,\mathrm m-\dfrac g2t^2$

The ball reaches the net when $x=12.0\,\rm m$ :

$12.0\,\mathrm m=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t\implies t=0.438\,\rm s$

At this time, the ball is at an altitude of

$2.34\,\mathrm m-\dfrac g2\left(0.438\,\mathrm s\right)^2=1.40\,\mathrm m$

which is 1.40 m - 0.900 m = 0.500 m above the net.

b. The change in angle gives the ball the new position functions

$x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t$

$y=2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)t-\dfrac g2t^2$

The ball reaches the net at time $t$ such that

$\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t=12.0\,\mathrm m\implies t=0.440\,\mathrm s$

at which point the ball's vertical position would be

$2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)\left(0.440\,\mathrm s\right)-\dfrac g2\left(0.440\,\mathrm s\right)^2=0.343\,\mathrm m$

so that the ball does not clear the net with 0.343 m - 0.900 m = -0.557 m.

5 0

3 years ago

What is the RICE regimen for injuries?

amid [387]

Answer:

<h2>R.I.C.E. stands for rest, ice, compression, and elevation</h2>

Explanation:

I hope this helps

7 0

3 years ago