What is (a) the x component and (b) the y component of the net electric field at the square's center

the orion nebula (at least the part we can see) is not very old (yet). while several hot, massive stars have had a chance to for

svetoff [14.1K]

One of the brightest nebulae in the night sky, the Orion Nebula may be seen with the unaided eye. The Trapezium is a young open cluster of four main stars in this magnitude 4 interstellar cloud of ionized atomic hydrogen.

<h3>What is the source of the Orion Nebula's crimson glow?</h3>

The hydrogen gas in the Orion Nebula, which is powered by radiation from young stars, gives off a crimson tint. The nebula's blue-violet regions are reflecting radiation from bright, blue-white O-type stars while the red areas are emitting light.
The Orion Nebula is one of many massive clouds of gas and dust in our Milky Way galaxy, say contemporary astronomers, and is one of the largest. It is approximately 1,300 light years away from Earth. This enormous hazy cocoon, which measures approximately 30 to 40 light-years in diameter, is generating potentially a thousand stars.

To learn more about Orion nebula refer to:

brainly.com/question/15575332

#SPJ4

3 0

11 months ago

A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a

ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω / t

(ω = √g/R)

∝ = $\frac{1}{t } \sqrt{\frac{g}{R} } }$

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = $\frac{2FR}{MR^2 + 2mr^2}$

Now we have ,

∝ = $\frac{1}{t} \sqrt{\frac{g}{R} }$ , ∝ = $\frac{2FR}{MR^2+2mr^2}$

equating both values

We have,

$F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }$

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

$F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N$

Each thruster has to applied a force of 294.5N in tangential direction

3 0

3 years ago

Calculate the force needed to accelerate a car of a mass 1000 kg by 3 m/s2

beks73 [17]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 1000 × 3

We have the final answer as

Hope this helps you

5 0

3 years ago

Can someone help me please? I’ve been trying to solve these questions all day.

qwelly [4]

#16

If we put a resistor in circuit it will slow the speed of current

Let's check ohms law

$\\ \rm\Rrightarrow \dfrac{V}{I}=R$

So if resistance is more current is less

#17

Again use ohms law

$\\ \rm\Rrightarrow V=IR$

$\\ \rm\Rrightarrow V\propto I$

Voltage must be increased

4 0

2 years ago

A long solenoid of radius 3 cm has 1100 turns per meter. If the solenoid carries a current of 1.5 A, then calculate the magnetic

Arada [10]

Answer:

The magnetic field at the center of the solenoid is 2.1 × 10⁻³ T

Explanation:

The magnetic field B at the center of the solenoid is given by

B = μ₀ni where μ₀ = permeability of free space = 4π × 10⁻⁷H/m, n = number of turns per unit length of the solenoid = 1100 turns per meter and i = current in the solenoid = 1.5 A.

So B = μ₀ni

= 4π × 10⁻⁷H/m × 1100 × 1.5 A

= 4π × 10⁻⁷H/m × 1650 A-turns/m

= 20734.5 × 10⁻⁷T

= 2.07345 × 10⁻³ T

≅ 2.1 × 10⁻³ T

So the magnetic field at the center of the solenoid is 2.1 × 10⁻³ T

3 0

3 years ago