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2 years ago

15

The voltage in the lines that carry electric power to homes is typically 2000 V. What is the required ratio of the loops in the

primary and secondary coils of the transformer to drop the voltage to 120 V?

2 answers:

Sav [38]2 years ago

7 0

Answer:

b

Explanation:

EDGE 2021

mariarad [96]2 years ago

6 0

Answer:

yurrrrr 221n54

Explanation:

yurrrrrrrrrrrrrrrrrrrrrr

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Una pelota de golf se lanza con una rapidez de 50 m/seg y un ángulo de elevación de 40º. Calcular:

miss Akunina [59]

Answer:

good morning you are you still

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2 years ago

Why is it a good idea to start with room temperature water in the calorimeter?

densk [106]

Explanation:

It is a good idea to start with room temperature water in the calorimeter because the room temperature water helps to determine the heating up/cooling down because of the environment as the experiment takes place. Because the calorimeter heat is the same as the heat of the water.

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3 years ago

Read 2 more answers

A horizontal force of 100 N is required to push a 50 kg crate across a factory floor at a constant speed. What is the accelerati

luda_lava [24]

Answer:

a = 2m/s^2

Explanation:

Force (F) = 100 N

Mass (m) = 50 kg

Here,

F = m×a

100 = 50 × a

a = 100÷50

a = 2m/s^2

Thus, the acceleration on the cart is a = 2m/s^2

-TheUnknownScientist

6 0

2 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago

A cord is attached to the box and run through a pulley directly above the box, so that the cord is vertical. The free end of the

Harman [31]

Answer:

The answer is given here would be a simplified equation, seeing as there are some missing variables in the question.

<u>F1 = T- 46, 674.656 gm/s² </u>

Explanation:

<em>Note: Once we have the mass of the second object and/or acceleration of the cord, we can solve for the force of the ground acting on the box.</em>

To calculate the force caused by gravity on the basic pulley system we use the following equation:

F2 = M2 x g; where g= gravitational acceleration (a constant equal to 9.8 m/s²). The mass M2 = 10.5 lb = 4762.72g

∴ F2 = 4762.72g x 9.8 m/s²

= 46, 674.656 gm/s² or 46, 674.656 N

But since this F2 is acting in a downlowrd direction, it would be negative.

Tension of the cord, T = Mass, x × acceleration. ( x is in the pulley diagram)

⇒ F1 = T - F2

<u>F1 = T- 46, 674.656 gm/s² </u>

4 0

2 years ago