All categories

3 years ago

What is in the center of our galaxy?

Physics

2 answers:

irakobra [83]3 years ago

8 0

Answer:

The Galactic center also known as "A supermassive black hole"

Explanation:

"Its the rotational center of the Milky way" -google

sattari [20]3 years ago

7 0

The answer would be the sun

You might be interested in

If 2 ma of current flow in your mp3 player, how long will it take for 1 c of charge to flow?

Mrrafil [7]

If 2 ma of current flow in your mp3 player, 8.3 mins or 500 sec will take for 1 c of charge to flow.

Electric Charge

Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge (commonly carried by protons and electrons respectively). Unlike charges attract one another while like charges repel one another. Neutral refers to an object that carries no net charge. Classical electrodynamics, the name given to an early understanding of how charged particles interact, is still accurate for issues that do not call for taking into account quantum phenomena.

The coulomb (C), which bears the name of French physicist Charles-Augustin de Coulomb, is the SI-derived unit of electric charge. The charge symbol, lowercase q, is frequently used.

To learn more about the electric charge refer here:

brainly.com/question/9194793

#SPJ4

7 0

2 years ago

What is the dimension symbols for energy

Alex_Xolod [135]

Answer:

The formula of energy is m $v^{2}$ . So, energy is M $L^{2}$ $T^{-2}$ . Since velocity = displacement/time.

Explanation:

If you have any questions feel free to ask in the comments - Mark

3 0

4 years ago

How do you find average velocity (average) from acceleration) and time (t)?

Tasya [4]

Average velocity is defined as the ratio in change in position to change in time,

v[ave] = ∆x/∆t

which on its own doesn't have anything to do with acceleration.

If acceleration is constant, the average velocity is the literal average of the initial and final velocities,

v[ave] = (v[final] + v[initial]) / 2

If this constant acceleration has magnitude a, the final velocity can be expressed in terms of the initial velocity by

v[final] = v[initial] + a*t

and plugging this into the previous equation gives

v[ave] = (v[initial] + a*t + v[initial])/2

v[ave] = v[initial] + 1/2*a*t

If the body in consideration is initially at rest, then

v[ave] = 1/2*a*t

which might be the relation you're looking for. But bear in mind the conditions I've underlined.

If acceleration is not constant and changes over time, so that the acceleration is some function of time a(t), then you can determine the velocity function v(t) by using the fundamental theorem of calculus. You need to know a particular velocity for some time to completely characterize v(t), though. For example, if you're given the initial velocity v[initial] = v(0), then

$\displaystyle v(t) = v(0) + \int_0^t a(u) \, du$

or if you know any other velocity for some time t₀ > 0,

$\displaystyle v(t) = v(t_0) + \int_{t_0}^t a(u) \, du$

8 0

3 years ago

Convert 9/4 hours into minutes

Eduardwww [97]

Am guessing;

9/4 × 60/1 =135

5 0

3 years ago

A long solenoid has a radius of 4.0 cm and has 800 turns/m. If the current in the solenoid is increasing at the rate of 3.0 A/s,

kolbaska11 [484]

Answer:

Explanation:

Given that,

Radius of solenoid R = 4cm = 0.04m

Turn per length is N/l = 800 turns/m

The rate at which current is increasing di/dt = 3 A/s

Induced electric field?

At r = 2.2cm=0.022m

µo = 4π × 10^-7 Wb/A•m

The magnetic field inside a solenoid is give as

B = µo•N•I

The value of electric field (E) can

only be a function of the distance r from the solenoid’s axis and it give as,

From gauss law

∮E•dA =qenc/εo

We can find the tangential component of the electric field from Faraday’s law

∮E•dl = −dΦB/dt

We choose the path to be a circle of radius r centered on the cylinder axis. Because all the requested radii are inside the solenoid, the flux-area is the entire πr² area within the loop.

E∮dl = −d/dt •(πr²B)

2πrE = −πr²dB/dt

2πrE = −πr² d/dt(µo•N•I)

2πrE = −πr² × µo•N•dI/dt

Divide both sides by 2πr

E =- ½ r•µo•N•dI/dt

Now, substituting the given data

E = -½ × 0.022 × 4π ×10^-7 × 800 × 3

E = —3.32 × 10^-5 V/m

E = —33.2 µV/m

The magnitude of the electric field at a point 2.2 cm from the solenoid axis is 33.2 µV/m

where the negative sign denotes counter-clockwise electric field when looking along the direction of the solenoid’s magnetic field.

3 0

3 years ago

Read 2 more answers