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3 years ago

QUICK PLZ I NEED TO TURN IT IN

Physics

2 answers:

Anni [7]3 years ago

8 0

The answer i got is 133.3 but i’m not completely sure . I got that because to get velocity i’m pretty sure it’s distance over time. So in between 8 and 10 minutes it’s 9 minutes. I pretty much just did 1200 divided by 9.

Serjik [45]3 years ago

7 0

Answer:800

Explanation:800 - 1600 = 800

Don’t really if this what you asked for but I tried.

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kykrilka [37]

Answer:

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Explanation:

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2 years ago

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What happens when you get a concussion?

madreJ [45]

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3 years ago

A group of cells working together make up a(n)

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2 years ago

A car radio draws 0.27 A of current in the autos 12-V electrical system. (a) How much electric power does the radio use?

Lostsunrise [7]

Answer:

(a) 3.24 w (b) 44.44 ohm

Explanation:

It is given that car draws 0.27 A current so current I = 0.27 A

The system has a voltage of 12 V

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(b) The resistance is defined as the ratio of voltage and current

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8 0

3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago