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Alenkasestr [34]
2 years ago
11

In the center of the Milky Way galaxy is a A

Physics
2 answers:
olchik [2.2K]2 years ago
5 0
The answer is the Galactic Center I think
LekaFEV [45]2 years ago
4 0

Answer:

a super massive blackhole

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A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0
3 years ago
A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas
Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, \theta=33^{\circ}

Time for which the player's foot is in contact with it, \Delta t = 5.1\times 10^{-2}\ s

Part A,

The x component of the soccer ball's change in momentum is given by :

\Delta p_x=mv\ cos\theta

\Delta p_x=0.425\times 15\ cos(33)

p_x=5.34\ kg-m/s

The y component of the soccer ball's change in momentum is given by :

\Delta p_y=mv\ sin\theta

\Delta p_y=0.425\times 15\ sin(33)

p_y=3.47\ kg-m/s

Hence, this is the required solution.

3 0
3 years ago
Your eyes see different wavelengths of light as different ________.
zaharov [31]

Answer:

Its color

Explanation: I got it right

5 0
3 years ago
A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica
Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2

\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2

26.01\times 40=26.082\times \omega _2

\omega _2=39.88 rad/s

3 0
3 years ago
In a double slit experiment, 450 nm light passes through two slits producing an interference pattern where the first bright frin
musickatia [10]

Answer:

1.3 x 10⁻⁴ m

Explanation:

\lambda = wavelength of the light = 450 nm = 450 x 10⁻⁹ m

n = order of the bright fringe = 1

θ = angle = 0.2°

d = separation between the slits

For bright fringe, Using the equation

d Sinθ = n \lambda

Inserting the values

d Sin0.2° = (1) (450 x 10⁻⁹)

d (0.003491) = (450 x 10⁻⁹)

d = 1.3 x 10⁻⁴ m

6 0
2 years ago
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