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3 years ago

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A car starts from rest and moves along the positive x-axis with a constant acceleration of 5 m/s/s for 8 seconds. It then contin

ues with constant velocity, i.e. the velocity it attains at 8 seconds for the next 5 seconds. What will be the distance covered by the car in the first 12 seconds of its motion?

2 answers:

boyakko [2]3 years ago

8 0

Answer is 340 plz believe

tino4ka555 [31]3 years ago

5 0

Answer:

340

Explanation:

Because you are moving at 5 m/s^2 for 8 seconds, you need to do some arithmetic to get the distance covered in the beginning. TO do this, you do 5 + 5*2 + 5*3 + 5*4 + 5*5 + 5*6 + 5*7 + 5*8, and once you have passed eight seconds, you need 4 more seconds with constant velocity, or 4(5*8). Your final equation would be 5+10+15+20+25+30+35+40+40+40+40+40. Hope this helps!

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You are fixing a transformer for a toy truck that uses an 8.0-V emf to run it. The primary coil of the transformer is broken; th

Art [367]

Answer:

a. The primary turns is 60 turns

b. The secondary voltage will be 360 volts.

Explanation:

Given data

secondary turns N2= 40 turns

primary turns N1= ?

primary voltage V1= 120 volts

secondary voltage V2= 8 volts

Applying the transformer formula which is

$\frac{N1}{N2} =\frac{V1}{V2}$

we can solve for N1 by substituting into the equation above

$\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60$

the primary turns is 60 turns

If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)

$\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360$

the secondary voltage will be 360 volts.

6 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

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Valentin [98]

B is correct but I’m not sure

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