Ask question

Ask question

All categories

3 years ago

5

What is a 3 letter (all lowercase) word that sits in the middle of the Milky Way? {Planets rotate and revolutionize around it...

}

1 answer:

Sergio [31]3 years ago

7 0

The sun is in the middle of the milky way and the planets in our solar system rotate around it

You might be interested in

If you want to go to a community college and transfer to a four-year university, what classes are required to transfer? How many

bagirrra123 [75]

Typically, most four-year institutions will require the following classes to transfer:

2 English composition courses.

1 Speech course.

1 Critical thinking course.

1 Physical science class.

1 Biological science class.

1 Health class.

2 Physical education courses.

Completion of college algebra

That's all I think

3 0

3 years ago

A thin plate 6 ft long and 3 ft wide is submerged and held stationary in a stream of water (T = 60°F) that has a velocity of 17

VashaNatasha [74]

A) In the case of the Boundary Thickness Layer we use the given formula,

$\delta = \frac{4.91x}{\sqrt{Re}}$

We know as well that,

Re = Número de Reynolds = $\frac{U*x}{\upsilon}$

Where,

U = velocity

$\upsilon$ = kinematic viscosity

For water, kinematic viscosity, $\upsilon = 1.21*10^{-5} ft^2 /s$

So, $500,000 = \frac{ 17x}{(1.21*10^{-5})}$

$x = 0.355 ft$

$d = \frac{4.91*0.355}{\sqrt {500000}}$

$d = 0.002465 ft = 0.029in$

B) For flat plate boundary layer. Given the Critical Reynolds Number.= 5*10^5 we know that is equal to Re above.

Thus, $x = 0.355 ft$

C. Wall shear stress,

$\tau = \mu*\sqrt{ U^3 / (2*\nu*x) }$

For water, dynamic viscosity, $\nu$ = 2.344*10^-5 lbf-s/ft^2

$\tau = 2.344*10^-5 \sqrt {17^3 / (2*1.21*10^{-5}*0.355)}$

$\tau = 0.5605 lbf/ft^2$

4 0

3 years ago

Your heart pumps blood at a pressure of 100 mmHg and flow speed of 60 cm/s. At your brain, the blood enters capillaries with suc

Savatey [412]

Answer:

1.28 m

Explanation:

Generally, pressure of fluid is given by

$P=\rho g h$ where g is acceleration due to gravity, h is the height and $\rho$ is the density

Considering that the pressure for mercury is same as for blood only that the height and density of fluid are different then

$\rho_b g h_b= \rho_m g h_m$

Since g is constant, then

$\rho_b h_b= \rho_m h_m$

Making $h_b$ the subject of the formula then

$h_b=\frac {\rho_m h_m}{\rho_b}$

Where subscripts m and b denote mercury and blood respectively

Assuming density of blood is 1060 Kg/m3, density of mercury as 13600 Kg/m3 and substituting height of mercury for 0.1 m then

$h_b=\frac {13600*0.1}{1060}=1.283018868 m \approx 1.28 m$

7 0

3 years ago

A 4.0 kg ball is traveling at 3.0 m/s and strikes a wall. The ball bounces off the wall with a velocity of 4.0 m/s in the opposi

trasher [3.6K]

Answer:

280 N

Explanation:

Applying Newton's third second law of motion,

F = m(v-u)/t................... Equation 1

Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.

Note: Let the direction of the initial velocity of the ball be positive

Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s

Substitute into equation 1

F = 4(-4-3)/0.1

F = 4(-7)/0.1

F = -28/0.1

F = -280 N.

Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball

3 0

3 years ago

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

4 0

3 years ago