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2 years ago

5

Calculate the angle θ between the radius-vector of the point and the positive x axis (measured counterclockwise from the positiv

e x axis, within the limits of −180◦ to +180◦ ). Answer in units of ◦

1 answer:

Y_Kistochka [10]2 years ago

6 0

The point obviously is in the 3rs quadrant

So

စ= tan^-1( y/x)-180

စ= -89.7°

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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

mihalych1998 [28]

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

6 0

3 years ago

Please just give me an answer, thanks.

Colt1911 [192]

#1

$\\ \rm\dashrightarrow P=VI\implies V=\dfrac{P}{I}$

$\\ \rm\dashrightarrow V=\dfrac{0.4\times 10^3}{33\times 10^{-3}}$

$\\ \rm\dashrightarrow V=0.012\times 10^{-6}$

$\\ \rm\dashrightarrow V=12mV$

#2

$\\ \rm\dashrightarrow R=\rho \dfrac{\ell}{A}$

$\\ \rm\dashrightarrow \rho =\dfrac{RA}{\ell}$

$\\ \rm\dashrightarrow \rho=\dfrac{86.3(0.4\times 10^6)}{69}$

$\\ \rm\dashrightarrow \rho=0.5\times 10^{6}$

It should be silicon

#3

Ohms law

$\\ \rm\dashrightarrow R=\dfrac{V}{I}$

$\\ \rm\dashrightarrow R=\dfrac{7(10^3)}{6(10^{-6})}$

$\\ \rm\dashrightarrow R=1.167\times 10^9\Omega$

7 0

2 years ago

A space probe is directly between two moons of a planet. If it is twice as far from moon A as it is from moon B, but the net for

Dennis_Churaev [7]

Answer:

c. Moon A is four times as massive as moon B

Explanation:

Let's assume the:

mass of the object = $m\,kilogram$

mass of the moon A = $M_A\,kilogram$

mass of the moon B = $M_B\,kilogram$

distance between the center of masses of the object and moon B = $r\,meters$

According to the given condition the object is twice as far from moon A as it is from moon B

∴distance between the center of masses of the object and moon B = $2r\,meters$

<u>As we know, gravitational force of attraction is given by:</u>

$F=G\frac{m_1.m_2}{r^2}$

<em>According to the condition</em>

Force on m due to $M_B=$ Force on m due to $M_A$

$G\frac{m.M_A}{(2r)^2} =G\frac{m.M_B}{(r)^2}$

$\frac{M_A}{4r^2} =\frac{M_B}{r^2}$

$M_A=4M_B$

3 0

3 years ago

The risk of earthquakes is high along the Pacific coast of the United States because a. There have been no earthquakes there lat

Ludmilka [50]

(D) That's where the Pacific and North American plates meet.

5 0

2 years ago

Solenoid constructed with 50 turns of wire and has inductance L. if we made another solenoid identical in length and cross secti

Vlad [161]

Answer:

It would have an inductance closest to 16 L.

Explanation:

Inductance for a one solenoid can be calculated with a formula following:

L=μ*N^2*A/l

Then, in this situation we are increasing the number of turns by 4 without any length change. First solenoid with 50 turn has inductance L which is:

L= μ*50^2*A/l=2500*μ*A/l

When we increase the number of turns by four, it will increase to:

L'=μ*200^2*A/l=40000*μ*A/l=16 L

6 0

3 years ago