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77julia77 [94]
3 years ago
5

Suppose there are two transformers between your house and the high-voltage transmission line that distributes the power. In addi

tion, assume that your house is the only one using electric power. At a substation the primary coil of a step-down transformer (turns ratio = 1:29) receives the voltage from the high-voltage transmission line. Because of your usage, a current of 53.3 mA exists in the primary coil of this transformer. The secondary coil is connected to the primary of another step-down transformer (turns ratio = 1:32) somewhere near your house, perhaps up on a telephone pole. The secondary coil of this transformer delivers a 240-V emf to your house. How much power is your house using? Remember that the current and voltage given in this problem are rms values.
Physics
1 answer:
storchak [24]3 years ago
8 0

Answer:240-v

Explanation:

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84. Three resitors each of value 30 respectively are connected in a parallel
kati45 [8]

Answer:

each resistor draws 1/3 of an amp or 0.33333 amps

Explanation:

V = I * R

V = 10 volts

R = 30 ohms

10 = I * 30       Divide by 30

10/30 = I

I = 0.33333

8 0
2 years ago
The electric field due to two point charges is found by: a) finding the stronger field. The net field will just be equal to the
alekssr [168]

Answer:

b)determining the electric field due to each charge and adding them together as vectors.

Explanation:

The electric Field is a vector quantity, in other words it has a magnitude and a direction. On the other hand, the electric field follows the law of superposition. The electric field produced by two elements is equal to the sum of the electric fields produced by each element when the other element is not present. in other words, the total electric field is solved determining the electric field due to each charge and adding them together as vectors.

7 0
3 years ago
Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the
Dafna11 [192]

\qquad\qquad\huge\underline{{\sf Answer}}♨

Heat capacity of body 1 :

\qquad \sf  \dashrightarrow \:m_1s_1

Heat capacity of body 2 :

\qquad \sf  \dashrightarrow \:m_2s_2

it's given that, the the head capacities of both the objects are equal. I.e

\qquad \sf  \dashrightarrow \:m_1s_1 = m_2s_2

\qquad \sf  \dashrightarrow \:m_1 =  \dfrac{m_2s_2}{s_1}

Now, consider specific heat of composite body be s'

According to given relation :

\qquad \sf  \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }

[ since, m_2s_2 = m_1s_1 ]

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 \cancel{m_2}s_2}{  \cancel{m_2}(\frac{s_2}{s_1} + 1)}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 s_2}{  (\frac{s_2 + s_1}{s_1} )}

\qquad \sf  \dashrightarrow \: s' =  \dfrac{2s_1s_2}{s_1 + s_2}

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

6 0
2 years ago
Read 2 more answers
A flea jumps straight up to a maximum height of 0.550 m . what is its initial velocity v0 as it leaves the ground?
Alexxx [7]
Since my givens are x = .550m [Vsub0] = unknown
 [Asubx] = =9.80
 
 [Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]) 

Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in

0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)

0 = [Vsub0x]^2 -10.78

10.78 = [Vsub0x]^2

Sqrt(10.78) = 3.28 m/s 


3 0
3 years ago
The drawing below shows two different types of pulley systems designed to lift a weight. In pulley system A, the end of the rope
nadya68 [22]
The answer would be in the chart or graph A is 1 B is 2
4 0
3 years ago
Read 2 more answers
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