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3 years ago

Please answer this!!!!! How does the light model help you explain how you can change the location of the shadow?

Physics

1 answer:

Alona [7]3 years ago

7 0

By moving the object, the light stays on the object, but moves with the object creating a shadow in a different area

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A 4.0 µC point charge and a 3.0 µC point charge are a distance L apart. Where should a third point charge be placed so that the

Rudik [331]

Answer:

Q must be placed at 0.53 L

Explanation:

Given data:

q_1 = 4.0 μC , q_2 = 3.0μC

Distance between charge is L

third charge q be placed at distance x cm from q1

The force by charge q_1 due to q is

$F1 = \frac{k q q_1}{x^2}$

$F1 = \frac{k q ( 4.0 μC )}{ x^2}$ ----1

The force by charge q_2 due to q is

$F2 = \frac{k q q_2}{(L-x)^2}$

$F2 = \frac{kq (3.0 μC)}{(L-x)^2}$ --2

we know that net electric force is equal to zero

F_1 = F_2

$\frac{k q ( 4.0 μC )}{x^2} =\frac{k q ( 3.0 μC )}{(l-x)^2}$

$\frac{4}{3}*(L-x)^2 = x^2$

$x = \sqrt{\frac{4}{3}*(L - x)$

$L-x = \frac{x}{1.15}$

$L = x + \frac{x}{1.15} = 1.86 x$

x = 0.53 L

Q must be placed at 0.53 L

3 0

3 years ago

The electric potential at a position located a distance of 20.7 mm from a positive point charge of 8.60×10-9C and 15.1 mm from a

max2010maxim [7]

Answer:

q2 = -4.35*10^-9C

Explanation:

In order to find the values of the second charge, you use the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$ (1)

V: electric potential = 1.14 kV = 1.14*10^3 kV

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1: charge 1 = 8.60*10^-9 C

q2: charge 2 = ?

r1: distance to the first charge = 20.7mm = 20.7*10^-3 m

r2: distance to the second charge = 15.1mm

You solve the equation (1) for q2, and replace the values of the other parameters:

$q_2=\frac{r_2}{k}[V-k\frac{q_1}{r_1}]=\frac{Vr_2}{k}-\frac{q_1r_2}{r_1}\\\\q_2=\frac{(1.14*10^3V)(15.1*10^{-3}m)}{8.98*10^9Nm^2/C^2}-\frac{(8.60*10^{-9}C)(15.1*10^{-3}m)}{20.7*10^{-3}m}\\\\q_2=-4.35*10^{-9}C$

The values of the second charge is -4.35*10^-9C

8 0

3 years ago

A child is pushing a shopping cart at a speed of 1.5 m/s how long will it take this child to push the cart down and I'll with th

maks197457 [2]

The answer to this question is 6.2 seconds

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3 years ago

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Which is the equivalent resistance of the circuit<br><br> shown below?

Misha Larkins [42]

1/Rt=1/R1+1/R2+1/R3, 1/Rt=1/3+1/12+1/4=2/3, Rt=equivalent resistance= 1.5 ohms

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3 years ago

You are walking up an icy slope. suddenly your feet slip, and you start to slide backward. will you slide at a constant speed, o

Aleks04 [339]

You will accelerate down the slope.

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3 years ago

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