Can anyone check if my table is correct or not?

A rod 14.0 cm long is uniformly charged and has a total charge of -22.0 μC. Determine the magnitude and direction of the net ele

lesya692 [45]

Explanation:

It is given that,

Length of the rod, l = 14 cm = 0.14 m

Charge on the rod, $q=-22\ \mu C=-22\times 10^{-6}\ C$

We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m

Electric field at the axis of the rod is given by :

$E=\dfrac{\lambda}{2\pi \epsilon_o z}$

Where

$\lambda$ is the linear charge density of the rod,

$\lambda=\dfrac{q}{l}=\dfrac{-22\times 10^{-6}\ C}{0.14\ m}=-0.00015\ C/m$

$E=\dfrac{-0.00015}{2\pi\times 8.85\times 10^{-12}\times 0.36}$

E = -7493170.57 N/C

or

$E=-7.49\times 10^6\ N/C$

Negative sign shows that the electric field is acting in inwards direction. Hence, this is the required solution.

4 0

3 years ago

According to the big bang theory, after the "bang," the universe remained dark until

Pepsi [2]

I believe the answer is 300,000 years

3 0

3 years ago

I am really struggling with this question because I can't find anything on aphelion and perihelion, it's not a topic we went ove

Hoochie [10]

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
69,816,900 km
0. 466 697 AU

 Distance at Perihelion (point in it's orbit that's closest to the sun):
46,001,200 km
0.307 499 AU 

Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is 1/2 of the orbital period.

Mercury's Orbital period = 87.9691 Earth days

1/2 (50%) of that is 43.9845 Earth days

The average of the aphelion and perihelion distances is

1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
1/2 ( 0.466697 + 0.307499) = 0.387 098 AU

This also happens to be 1/2 of the major axis of the elliptical orbit.

3 0

4 years ago

Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black

arlik [135]

Answer:

$833.4801043*10^6N$ on ear that is closer to Black hole.

$13.83803929*10^6N$ On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

$F =\frac{m_{1}*m_{2} }{r^2} G$