The dimensions of the disk of our milky way galaxy are select one:

This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e

Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

$P = \frac{1}{2} \rho v^3 A C_p$

where $\rho = 1.2 kg/m^3$ is the air density, v = 8.89 m/s is the wind speed, A is the swept area and $C_p = 0.45$ is the efficiency

$10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45$

$10000 = 190A$

$A = 10000 / 190 = 52.7 m^2$

The swept area is a circle with radius r being the blade length

$\pi r^2 = A = 52.7$

$r^2 = 52.7 / \pi = 16.79$

$r = \sqrt{16.79} = 4.1 m$

4 0

3 years ago

The line of latitude that is zero degrees latitude is the

Nuetrik [128]

Answer:

Equator

Explanation:

The equator is the line of latitude that separates the earth into the northern and southern hemisphere.Its magnitude is 0 degrees latitude. The equator and other lines of latitude are parallel to each other. The lines of longitude which is 0 degrees is known as the Greenwich meridian. The longitudinal lines however are not parallel as they converge at the poles.

7 0

3 years ago

An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq

Andru [333]

Answer:

f = 409 Hz

Explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone, $f_2 = 1227 Hz$

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

$f_2=\dfrac{3v}{2l}$

v is speed of sound

Let f is the fundamental frequency. It is given by :

$f=\dfrac{v}{2l}$

The relation between f and f₂ can be written as :

$f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz$

So, the fundamental frequency of the pipe is 409 Hz.

6 0

2 years ago

Question 1 (1 point)

Sonbull [250]

Answer:

Solid objects will deform when adequate loads are applied to them; if the material is elastic, the object will return to its initial shape and size after removal. This is in contrast to plasticity, in which the object fails to do so and instead remains in its deformed state.

Explanation:

4 0

2 years ago

Read 2 more answers

A spherical ball of charged particles has a uniform charge density. In terms of the ball's radius R, at what radial distances in

eimsori [14]

Answer:

Electric field at radius r inside the solid sphere is

Electric field at radius r between inner radius and outer radius of the shell is

E=0 N/C

Explanation:Given

The radius of the solid sphere is

The charge on the solid sphere is

The inner radius of the shell is

The outer radius of the shell is

The total charge on the shell is

PART(A)

The magnitude of electric field at radius r where \\The volumetric charge density of the solid sphere will be

The charge enclosed by the radius r inside the solid sphere is

rho=

According to gauss law

PART(B)

The electric field at radius r where

The shell is conducting so due to induction of charge

The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell

So,

The charge on the inner surface of the shell is

Due to conservation of the charge on the shell

The charge accumulated on the outer surface of the shell is

The charge enclosed by the radius r where

According to gauss law

Read more on Brainly.com - brainly.com/question/13242041#readmore

3 0

2 years ago