A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.9 cm ( 0.589 m) and the flow speed of the petroleum is 12.1 m/s. At the refinery, the petroleum flows at 6.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer 1
Answer:

Answer:

Explanation:

The volume rate of flow = a x v where a is cross sectional area of pipe and v is velocity of flow

putting the values

π x .2945² x 12.1

= 3.3  m³ /s

To know the pipe's diameter at the refinery we shall apply the following formula

a₁ v₁ = a₂ v₂

a₁ v₁ and  a₂ v₂ are volume rate of flow of liquid which will be constant .

3.3 = a₂ x 6.29

a₂ = .5246 m³

π x r² = .5246

r = .4087 m

= 40.87 cm

diameter

= 81.74 cm


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As a science project, you drop a watermelon off the top of the Empire State Building. 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a constant speed of 30 m/s. A) How much time passes before the watermelon has the same velocity? B) How fast is the watermelon going when it passes Superman?C) How fast is the watermelon traveling when it hits the ground?

Answers

Answer:

3.06 seconds time passes before the watermelon has the same velocity

watermelon going at speed 59.9 m/s

watermelon traveling when it hits the ground at speed is 79.19 m/s

Explanation:

given data

height = 320 m

speed = 30 m/s

to find out

How much time passes before the watermelon has the same velocity and How fast is the watermelon going and How fast is the watermelon traveling

solution

we will use here equation of motion that is

v = u + at    ....................1

here v is velocity 30 m/s and u is initial speed i.e zero and a is acceleration i.e 9.8 m/s²

put the value and find time t

30 = 0 + 9.8 (t)

t = 3.06 s

so 3.06 seconds time passes before the watermelon has the same velocity

and

we know superman cover distance is = velocity × time

so distance = 30 × t

and distance formula for watermelon is

distance = ut + 0.5×a×t²    .............2

here u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and h is 30 × t

30 × t = 0 + 0.5×9.8×t²

t = 6.12 s

so  by equation 1

v = u + at

v = 0 + 9.8 ( 6.12)

v = 59.9 m/s

so watermelon going at speed 59.9 m/s

and

watermelon traveling speed formula is by equation of motion

v² - u² = 2as      ......................3

here v is speed and u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and s is distance i.e 320 m

v² - 0 = 2(9.8) 320

v = 79.19 m/s

so watermelon traveling when it hits the ground at speed is 79.19 m/s

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?

Answers

Answer:

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

\Delta x = (\lambda L)/(d)\n\n\lambda = (\Delta x d)/(L)

where,

λ = wavelength = ?

L = distance between screen and slits = 8.61 m

d = slit spacing = 1.09 mm = 0.00109 m

Δx = distance between consecutive bright fringes = (4.53\ cm)/(10) = 0.00453 m

Therefore,

\lambda = ((0.00453\ m)(0.00109\ m))/(8.61\ m)

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Draw a ray diagram for an object placed more than two focal lengths in front of a converging lens.

Answers

Answer:

Explanation:

There is a convex lens M N is placed. An object AB is placed at a distance more than two focal lengths of the lens.

A ray of light is starting from point A and parallel to the principal axis, then after refraction it goes from the focus.

Another ray which goes through the optical centre of the lens becomes undeviated after refraction.

The two refracted rays meet at the point A', So A'B is the image of AB.

The nature of image is real, inverted and diminished.

(a) A 2.00-µF capacitor is connected to a 18.0-V battery. How much energy is stored in the capacitor?

Answers

Answer:

1.8 x 10⁻⁵J

Explanation:

The energy (E) stored in a capacitor of capacitance, C,  when a voltage, V, is supplied is given by;

E = (1)/(2) x C x V²              -------------------(i)

Now, from the question;

C = 2.00μF = 2.00 x 10⁻⁶F

V = 18.0V

Substitute these values into equation (i) as follows;

E = (1)/(2) x 2.00 x 10⁻⁶ x 18.0

E = 1.8 x 10⁻⁵J

Therefore, the quantity of energy stored in the capacitor is 1.8 x 10⁻⁵J

Answer:

1.8 x 10⁻⁵J.

Explanation:

E = 1/2 x C x V²

Where,

E = energy stored in a capacitor

C = capaciitance

V = Voltage

From the question, given:

C = 2.00μF

= 2.00 x 10⁻⁶F

V = 18.0V

E = 1/2 x 2.00 x 10⁻⁶ x 18.0

= 1.8 x 10⁻⁵J.

Of the four most important pathways by which stress affects health, the first one to occuris usually related to physiology

Answers

Answer: Physiologic response to fear is very similar to that of PTSD and stress. Fear is accompanied by increased heart rate due to the release of adrenaline, sympathetic nervous system is aroused. The release of adrenaline also causes increased sweating, pulse and blood pressure. In line with this, the parasympathetic nervous system experiences reduced activity such as decrease in digestion.

Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done by the force

Answers

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

\Rightarrow a=(F)/(m)

Hence acceleration of the block is given by:

\Rightarrow a=(25)/(2)=12.5m/s^2

Displacement of the object is given by:

\Rightarrow S=ut+(1)/(2)at^2

Substitute the values

\Rightarrow S=0*3+(1)/(2)(12.5)3^2

\Rightarrow S=56.25 m

Now work done is given by:

 W=F.S

W = 25×56.25

W =1406.25 J