All categories

2 years ago

The unusually bright centers found in some galaxies are called ______.

Physics

1 answer:

Nimfa-mama [501]2 years ago

4 0

Active Galactic Nuclei.

You might be interested in

You are looking down on a N = 9 turn coil in a magnetic field B = 0.5 T which points directly down into the screen. If the diame

Novay_Z [31]

Answer:

The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.

Explanation:

Given that,

Number of turns = 9

Magnetic field = 0.5 T

Diameter = 3 cm

Time t = 0.14 s

We need to calculate the flux

Using formula of flux

$\phi=NAB$

Put the value into the formula

$\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5$

$\phi=0.003180$

We need to calculate the emf

Using formula of emf

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{0.003180}{0.14}$

$\epsilon =-0.000227\ V$

$\epsilon=-2.27\times10^{-4}\ V$

Negative sign shows the direction of current.

Hence, The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.

8 0

3 years ago

A ball is dropped from a height h and falls the last half of its distance in 4 seconds. How long does the ball fall? From what h

dlinn [17]

Answer:

How long does the ball fall is t_2 = 13.66 (s).

From what height is the ball originally dropped is h= 913.90 (m).

Explanation:

8 0

2 years ago

So lost on this one number 1 ?

kirza4 [7]

Read each statement, one at a time, and form the picture in your mind.

Statement #1 should make you feel all warm and fuzzy inside.

Statements #2, #3, and #4 should make your red flag wave on the inside, and should make you giggle on the outside when you realize how absurd each of those is.

5 0

2 years ago

An atom that no longer has a beutral charge is called a(n)____.

Simora [160]

An atom that no longer has a neutral charge is called an ion

8 0

2 years ago

ASAP please!! The force vectors on an aircraft are as shown. Find the net (resultant force).

ioda

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

$W_D=3600 N\cdot\sin 27^\circ=1634N$

acting in the direction of the Drag, and the plane-perpendicular component is:

$W_L=-3600N\cdot\cos 27^\circ=-3208N$

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

$F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N$

The magnitude of the net force is:

$|F|=\sqrt{5366^2+829^2}N = 5430N$

6 0

3 years ago