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irina1246 [14]
3 years ago
14

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm

. Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.
Physics
1 answer:
mihalych1998 [28]3 years ago
6 0

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?

Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.  

(a) What is the electric potential at point a due to q1 and q2?  

(b) What is the electric potential at point b?

(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?

Answer:

a) the potential is zero at the center .

Explanation:

a) since the two equal-magnitude and oppositely charged particles are equidistant

b)(b) Electric potential at point b, v = Σ kQ/r

r = 5cm = 0.05m

k = 8.99*10^9 N·m²/C²

Q = -2 microcoulomb

v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m

v =  -105 324 V

c)workdone = charge * potential

work = -6.00µC * -105324V

work = 0.632 J

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metal

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metal is an excellent conductor for heat and electricity therefore making it perfect to experiment with different temperatures.

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A race horse can run a mile race in just under 2 minutes. Is it possible for
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Yes.

Explanation:

A kilometer is less than a mile, therefore if a horse can finish one mile in less than 2 minutes then it can certainly do a kilometer in less than two minutes.

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2 years ago
In an electromagnetics lab, you are studying two coils which have a mutual inductance of M=300 mH. Suppose that the current in t
Firdavs [7]

Answer:

Explanation:

Given that,

The mutual inductance of the two coils is

M = 300mH = 300 × 10^-3 H

M = 0.3 H

Current increase in the coil from 2.8A to 10A

∆I = I_2 - I_1 = 10 - 2.8

∆I = 7.2 A

Within the time 300ms

t = 300ms = 300 × 10^-3

t = 0.3s

Second Coil resistance

R_2 = 0.4 ohms

We want to find the current in the second coil,

The same induced EMF is in both coils, so let find the EMF,

From faradays law

ε = Mdi/dt

ε = M•∆I / ∆t

ε = 0.3 × 7.2 / 0.3

ε = 7.2 Volts

Now, this is the voltage across both coils,

Applying ohms law to the second coil, V=IR

ε = I_2•R_2

0.72 = I_2 • 0.4

I_2 = 0.72 / 0.4

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7 0
3 years ago
A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 95 km/h to zero is 55 m.
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Answer:

-0.64525g

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 95 km/h

v = Final velocity = 0 km/h

s = Displacement

a = Acceleration

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2

Converting to m/s²

a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2

g = Acceleration due to gravity = 9.81 m/s²

Dividing both the accelerations, we get

\frac{a}{g}=\frac{-6.33}{9.81}=-0.64525\\\Rightarrow a=-0.64525g

Hence, acceleration of the car is -0.64525g

8 0
4 years ago
Diffusion and osmosis are forms of passive transport.<br><br> True<br> False
pishuonlain [190]

Answer:

True. Diffusion and osmosis are forms of passive transport.

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In diffusion, particles move from an area of higher concentration to one of lower concentration until equilibrium is reached.

In osmosis, a semipermeable membrane is present, so only the solvent molecules are free to move to equalize concentration.

8 0
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