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3 years ago

The center of the Milky Way most likely contains

Physics

1 answer:

Naya [18.7K]3 years ago

3 0

Answer:

The center of the Milky Way most likely contains a supermassive black hole.

Explanation:

Because it is an eleptical galaxy, it has a little rotation to it but not enough to flatten out so the center will contain a supermassive black hole.

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Write down the symbols of nitrogen and neon?

statuscvo [17]

Answer:

Nitrogen is N, neon is Ne

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2 years ago

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The repulsive force between two protons has a magnitude of 2.00 N. What is the distance between them?

Aloiza [94]

Answer:

The answer to your question is letter A. r = 1.07 x 10⁻¹⁴ m

Explanation:

Data

F = 2 N

d = ?

q = 1.6 x 10 ⁻¹⁹ C

k = 8.987 Nm²/C²

Formula

$F = K\frac{q1q2}{r^{2}}$

Solve for r

$r = \sqrt{\frac{kq1q2}{F}}$

Substitution

$r = \sqrt{\frac{8.987 x 10^{9}x1.6 x 10^{-19} x 1.6 x 10x^{-19}}{2}}$

Simplification

r = $\sqrt{\frac{2.3 x 10^{-28}}{2}}$

r = $\sqrt{1.15 x 10^{-24}}$

Result

r = 1.07 x 10⁻¹⁴ m

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3 years ago

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What is the equivalent resistance of the circuit?

mezya [45]

Answer:

I=VRS=9V90Ω=0.1A

Explanation:

The equivalent resistance is the algebraic sum of the resistances (Equation 10.3. 2): RS=R1+R2+R3+R4+R5=20Ω+20Ω+20Ω+20Ω+10Ω=90Ω. The current through the circuit is the same for each resistor in a series circuit and is equal to the applied voltage divided by the equivalent resistance

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3 years ago

An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.

Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }$

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }$

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }$

$B = \boxed{7.07 *10^{-10} T}$

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

$B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}$