All categories

3 years ago

HELP PLS I JUST NEED THE ANSWER PLSSSSSSSSSSSSSSSSSSSSSS ASAP HELPPPP

Physics

1 answer:

Artist 52 [7]3 years ago

7 0

Answer:

Explanation:

Pooh Shiesty, that's my dawg

But Pooh, you know I'm really shiesty

You told all them O.T. ni ggas that you really slide

Tell the truth about your gang, they really dyin'

I got my own fire, don't need security in the club

All that woofin' on the net , I thought you was a thug

They ain't got nowhere to go, I sh0t up everywhere they was

Yeah, you know who took that - from you

Come get it back in blood

You might be interested in

Which kind of pressure prevents stars of extremely large mass from forming?

elena-14-01-66 [18.8K]

Answer:

Radiation pressure

Explanation:

Radiation pressure is gotten from the light source. The radiation pressure has a set mass limit of 150 solar masses. Any mass from 150 solar masses or below supports the formation of the stars.

However when the mass exceeds 150 solar masses the force of gravity is unable to hold the masses together and the excess mass

Is usually blown away after which the required mass is then used in the formation.

3 0

4 years ago

Read 2 more answers

Pani-rosa [81]

The total displacement is 4.0 m east.

4 0

3 years ago

How does an impulse propagate down the axon

kolezko [41]

When action potentials reach the end of the axon, they stimulate opening of Ca2+ channels, causing a release of neurotransmitters to the post-synaptic cell. How does and impulse propagate down the axon? The stimulus causes a start of the action potential and it moves down the axon without the ions moving down

5 0

3 years ago

A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.

Contact [7]

Answer:

A) $V_A = 11.93~V$

B) The vector definition of E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

$V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}$

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

$r = \sqrt{x^2+4^2}$

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

$dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}$

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

$V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx$

By using an integral table, this can be calculated:

$V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V$

B) The electric field can be found by a similar approach, but a different formula:

$\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\^r$

Let's apply this formula to the infinitesimal part we have chosen.

$dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)$

By the geometry sine and cosine terms can be found:

$\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}$

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

$E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx = 2.41(\^y)$

So, the final E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

The magnitude of the E-field is

E = 2.66 N/m

6 0

3 years ago

Why do planets revolve around the sun?

stepladder [879]

Answer:

The gravity of the Sun keeps the planets in their orbits. They stay in their orbits because there is no other force in the Solar System which can stop them.

Explanation:

5 0

3 years ago