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AlekseyPX
2 years ago
7

Differentiate between step up and step down transformer​

Physics
1 answer:
Sav [38]2 years ago
5 0

Answer:

The main difference between the step-up and step-down transformer is that the step-up transformer increases the output voltage, while the step-down transformer reduces the output voltage.

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That's a good question.
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3 years ago
Read 2 more answers
Calculate the energy gained by an ice block in the following experiment.
MAXImum [283]

The heat is exchanged when two different temperature objects come in contact. The energy gained by an ice block is 2.3 Joules.

<h3>What is temperature?</h3>

Temperature is the degree of hotness and coldness of the object.

A 7g block of ice was added to a coffee cup full of 103.4 grams of water. The water had an initial temperature T₁ = 24.5 C and a final temperature T₂ = 19.2 C after all the ice had melted.

Heat lost by water = Heat gained by ice

Qgain = ms(T₂ -T₁ )

Substituting the value for mass of water m =103.4 g= 0.1034 kg , specific heat of water s = 4.18 kJ/kg and temperature values, we get

Qgain = 0.1034 x 4.18 x (24.5 - 19.2)

Qgain = 2.3 Joules

Thus, the energy gained by an ice block is 2.3 Joules.

Learn more about temperature.

brainly.com/question/20459283

#SPJ1

7 0
2 years ago
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Do thermohaline currents flow vertically or horizontally<br><br> A. Vertically<br> B. Horizontally
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7 0
3 years ago
A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,
kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}

so that the <em>x</em>-component of \vec v_{\rm ave} is

\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}

and its <em>y</em>-component is

\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0
3 years ago
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