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elixir [45]
3 years ago
9

All the mass we ever encounter is positive - nothing weighs less than zero. What makes us think charge is different that there a

re both positive AND negative charges?
Physics
1 answer:
egoroff_w [7]3 years ago
4 0

Based on the number of positive and negative charges

<u>Explanation</u>:

  1. In a positive charge particle, the  number of proton is higher than the number of electron. Due to this fact alone, we can come to the conclusion that positive charge consists of more number of protons than electrons.
  2. In a negative charge particle, the  number of electron is higher than the number of proton. Due to this fact alone, we can come to the conclusion that negative charge consists of more number of electrons than protons.
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dem82 [27]
Acceleration is a change in *speed* over time. In this case, the speed of the car increased by 90 km/hr in 6 s, giving it a rate of 90 km/hr/6s, or 15 km/hr/s. We’re asked for the acceleration in m/s^2, though, so we’ll need to do a few conversions to get our units straight.

There are 1000 m in 1 km, 60 min, or 60 * 60 = 3600 s in 1 hr, so we can change our rate to:

(15 x 1000)m/3600s/s, or (15 x 1000)m/3600 s^2

We can reduce this to:

(15 x 10)m/36 s^2 = 150 m/36 s^2

Which, dividing numerator and denominator by 36, gets us a final answer of roughly 4.17 m/s^2
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4 years ago
Answer <br> A<br> B<br> C<br> D and <br> E
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Answer:

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Explanation:

4 0
3 years ago
The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm
Alika [10]

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force,  F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by F=kx

So 63.5=k\times 0.0531

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J

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3 years ago
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3 years ago
Read 2 more answers
A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of
enot [183]

Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

\frac{n_1}{n_2}=\frac{V_1}{V_2}

In the old transformer the ratio of voltages was:

\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1

In the new transformer the ratio of voltages is:

\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1

If the old current output was 425 kV, the ratio of current was:

\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{425}= 0.028\\\\I_2=0.028I_1

Then, the ratio of the new output over the old output is:

\frac{I_{2new}}{I_{2old}} =\frac{0.024\cdot I_1}{0.028\cdot I_1}= 0.86

The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).

c) The power loss is expressed as:

P_L=I^2\cdot R

Then, the ratio of losses is (R is constant for both power losses):

\frac{P_n}{P_o} =\frac{I_n^2R}{I_o^2R} =(\frac{I_n}{I_o} )^2=0.86^2=0.74

The losses in the new line are 74% the losses in the old line.

7 0
3 years ago
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