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Roman55 [17]
4 years ago
5

In the diagram, q1 = +1.39*10^-9 C and q2 = +3.22*10^-9 C. The electric field at point P is zero. What is the distance from P to

q2?

Physics
1 answer:
Sphinxa [80]4 years ago
6 0

Answer:

0.388

Explanation:

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A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is
vova2212 [387]

solution:

consider the following data\\
length of slicon bar with circular cross section is 4cm or 0.04m\\
at room temperature resistance of the slicon bar is 270\Omega \\
represent the resistance in mathematical from\\
r=p\frac{1}{A}---1\\
where r is resistance and l is the length \\
A is cross sectional area\\
it is clear that resistivity of the silicon meterial is 6.4\times^2 \Omega.m\\
substitute 6.4\times10^2 for p,270\Omega for R and 0.04m for l i equation (1).\\270=(604\times10^2)\frac{0.04}{A}\\
rewrite the equation\\
a=(6.4\times10^2)\frac{(0.04)}{270}\\
=0.9481m^2\\
write the formula for the circular cross sectional area of silicon bar.\\
A=\pi r^2\\
substitute 0.9481 for A in the above equation\\
\pi r^2=0.9481
r^2=\frac{0.9481}{3.14},since \pi =3.14\\
0.30194\\
further simplified\\
r^2=0.30194\\
\sqrt{0.30194}\\
\cong 0.1509m\\
\cong 150.1mm

7 0
4 years ago
5. Alex goes cruising on his dirt bike. He rides 700 m north, 300 m east, 400 m north, 600 m west, 1200 m south 300 m east and f
Gala2k [10]

Answer:

Explanation:

i hope your mind is clear your full answer in image

please see in image

5 0
3 years ago
A bicycle has a momentum of 36 kg • m/s and a velocity of 4 m/s. What is the mass of the bicycle? 9 kg 32 kg 40 kg 144 kg
igomit [66]

Answer:

9 kg

Explanation:

Momentum is mass times velocity.

p = mv

36 kgm/s = m (4 m/s)

m = 9 kg

3 0
3 years ago
Read 2 more answers
A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Su
aniked [119]

Answer:

The planet´s orbital period will be one-half Earth´s orbital period.

Explanation:

The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.

At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:

ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:

Fg = Fc ⇒   G*mp*ms / r² = mp*(2*π/T)²*r

Solving for T²:

T² = (2*π)²*r³ / G*ms (1)

For the planet orbiting the sun in Andromeda, we have:

Ta² = (2*π)*r³ / G*4*ms (2)

As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:

(Te / Ta)² =  4  ⇒ Te / Ta = 2 ⇒ Ta = Te/2

So, The planet's orbital period will be one-half Earth's orbital period.

7 0
3 years ago
Which part of the following graph shows the object going at the fastest speed?
Alex73 [517]

Answer:

more point plzzzzz... eeeeeeeee

4 0
3 years ago
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