Once out in space, the space shuttle needs to ________________ to keep going in the same direction at the same speed.

Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In

OLEGan [10]

Answer:

discharge rate (Q) = 0.2005 m^{3} / s

Explanation:

if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

solution:

time = 1 s

elevation of point 1 (z1) = 10 m

elevation of point 2 (z2) = 2 m

elevation of point 3 (z3) = 2 m

cross section area of point 2 = 4.8 x 10^{2} m

cross section area of point 3 = 1.6 x 10^{2} m

g

acceleration due to gravity (g) = 9.8 m/s^{2}

find the discharge rate at point 3 which is the exit pipe.

discharge rate (Q) = A3 x V3

where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below

$\frac{P1}{ρg} + \frac{V1^{2} }{2g} + Z1 = \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3$

pressure at point 1 (P1) is the same as pressure at point 3 (P3), and at point 1, the velocity (V1) = 0. therefore the equation now becomes

$\frac{P1}{ρg} + Z1 = \frac{P1}{ρg} + \frac{V3^{2} }{2g} + Z3$

$Z1 = \frac{V3^{2} }{2g} + Z3$

V3 = $\sqrt{2g(Z1-Z3)}$

V3 = $\sqrt{2 x 9.8 x (10 - 3)}$

V3 = 12.53 m/s

discharge rate (Q) = A3 x V3 = 1.6 x 10^{-2} x 12.53

discharge rate (Q) = 0.2005 m^{3} / s

8 0

3 years ago

An architect is redesigning a rectangular room on the blueprints of a house. She decides to double the width of the room, increa

Vladimir79 [104]

Answer:3.6

Explanation:

Given

Architect decide to double the width

length increases by 50 percent

height increases by 20 percent

Let L,B,H be length,height and width of rectangular room

L'=1.5L

B'=2B

H'=1.2H

therefore new volume V'=L'\times B'\times H'

$V'=1.5L\times 2B\times 1.2H$

V'=3.6LBH

new volume is 3.6 times the original

8 0

3 years ago

What type of spectrum is emitted by a low-density luminous cloud of hot gases?

hoa [83]

The answer is Emission

5 0

3 years ago

A ball rolls off the edge of a table with a fairly large horizontal velocity. Which of the following statements are true? (Selec

Degger [83]

Answer:

A.The vertical velocity is constantly increasing as the ball falls.

B.The horizontal velocity does not noticeably change as the ball falls.

G.The horizontal velocity does not affect how long it will take the ball to fall to the floor.

H.The velocity vector of the ball changes as it travels through the air.

Explanation:

As the ball is projected horizontally so here the vertical component of the velocity is zero

So the time to reach the ground is given as

$H = \frac{1}{2} gt^2$

so we will have

$t = \sqrt{\frac{2H}{g}}$

so this is the same time as the ball is dropped from H height

Since there is no force in horizontal direction so its horizontal velocity will always remain constant while vertical velocity will change at constant rate which is equal to acceleration due to gravity.

So overall the velocity vector will change due to net acceleration g

4 0

3 years ago

In a binary-star system that produces a nova, the white dwarf pulls matter from the companion star. The matter forms an accretio

Studentka2010 [4]

As a head-up, it is important to notice that a white dwarf only shines thanks to the stored energy and light, because a white dwarf doesn't have any hydrogen left to perform nuclear fusion.

Now the process:

First, the white dwarf accumulates all the extracted matter from its companion, onto its own surface. This extra matter increases the white dwarf's temperature and density.

After a while, the star reaches about 10 million K, so nuclear fusion can begin. The hydrogen that has been "stolen" from the other star and accumulated in the white dwarf's surface it's used for the fusion, dramatically increasing the star's brightness for a short time, causing what we know as a Nova.

As this fuel its quickly burnt out or blown into space, the star goes back to its natural white dwarf state. Since the white dwarf nor the companion star are destroyed in this process, it can happen countless of times during their lifespan.

4 0

3 years ago