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kaheart [24]
3 years ago
15

A transformer for a laptop computer converts a 120-V input to a 24-V output. If the output current for the transformer is 1.8 A,

find the input current.
Physics
1 answer:
astra-53 [7]3 years ago
3 0
To find for the input current, we use ratio and proportion concept. We use the equation:

I(input)/ I(output) = V(output)/ V(input)

I(input) / 1.8 A = 24 V / 120 V
I(input) = 0.36 A

Hope this answers the question. Have a nice day. Feel free to ask more questions.
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What happens to the resistance of a wire as it gets wider
Alchen [17]

Answer:

it snaps

Explanation:

the more force you put on it, the wired out it gets than it snaps. I think

5 0
3 years ago
A gamma ray is pure __________. <br><br><br><br><img src="https://tex.z-dn.net/?f=%20u%5C%5C%20r%5C%5C%20cursed%5C%5C%20my%5C%5C
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3 years ago
What is the angle θ between vectors A⃗ and B⃗ if A⃗ =4ı^−4ȷ^ and B⃗ =−5ı^+7ȷ^?
ololo11 [35]

The characteristics of the scalar product allows to find the angle between the two vectors is:

  • The angle θ = 170º

The scalar product is the product between two vectors whose result is a scalar.

            A . B = |A|  |B| cos θ

Where A and B are the vectors, |A| and |B| are the modules of the vectors and θ at the angle between them.

The vector is given in Cartesian coordinates and the unit vectors in these coordinates are perpendicular.

            i.i = j.j = 1

            i.j = 0

            A . B = (4 i - 4j). * -5 i + 7j)

            A . B = - 4 5 - 4 7

            A. B = -48

We look for the modulus of each vector.

           |A| = \sqrt{x^2 +y^2 }

           |A| = \sqrt{4^2 + 4^2}  

           |A| = 4 √2

          |B| = \sqrt{5^2 +7^2}

          |B| = 8.60

We substitute.

            -48 = 4√2  8.60  cos θ

            -48 = 48.66 cos θ

            θ = cos⁻¹   \frac{-48}{48.664}  

            θ = 170º

In conclusion using the dot product we can find the angle between the two vectors is:

  • the angle θ = 170º

Learn more about the scalar product here:  brainly.com/question/1550649

8 0
3 years ago
An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi
Svetllana [295]

Answer:

a

P_G  = 14.03 \  psig  

b

h_m =   0.148 \  m

Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is P_{m} =  15.5 \  psig  =  15.5 *  6894.76 =  106868.78 \ N/m^2

The level of mercury is h  =  950 \ mm  =  0.950 \  m

The drop in the mercury level at the visible arm is d =  39.0 =  0.039 \  m

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

P_g  =  P_m  =  g *  \delta h  * \rho

Here \rho is the density of mercury with value \rho = 13.6 *10^{3} kg/m^3

and \delta h is the difference in the level of gas in arm one and two

So

\delta h  =  \frac{106868.78}{  13.6 *10^{3} *  9.8 }

\delta h  = 0.802 \  m

Generally the height of the mercury at the arm connected to the pipe is mathematically represented as

h_m =   0.950 -  0.802

=> h_m =   0.148 \  m

Generally from manometry principle we have that

P_G + \rho * g  * d   -  \rho *  g  * [h - (h_m + d)] = 0

Here P_G is the pressure of the gas

P_G +13.6 *10^{3} * 9.8  * 0.039    -  13.6 *10^{3}  *  9.8  * [0.950 - (0.148 + 0.039)] = 0

P_G  =  9.6724 04 *10^{4} \  N/m^2

converting to  psig

P_G  = \frac{ 9.6724 04 *10^{4} }{6894.76}

P_G  = 14.03 \  psig

6 0
3 years ago
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