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4 years ago

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Two pipes move the same amount of ideal fluid in the same amount of time. One pipe has a 2 in. diameter; the other has a 3 in. d

iameter. In which pipe is the fluid pressure higher, assuming that all other conditions are the same?a) 3-in. pipe b) 2-in. pipe c) same pressure in both d) cannot be determined for an ideal fluid

1 answer:

KATRIN_1 [288]4 years ago

6 0

Answer:

a) 3-in. pipe

Explanation:

Given that

Fluid flow is in same amount in the same time it means that volume flow rate is same for the pipes

Volume flow rate

Q = A V

A=Area ,V=Velocity

$A=\dfrac{\pi}{4}d^2$

If diameter d is more then the velocity will be less for same volume flow rate .We also Know that if pressure is more then the velocity will be less.

The second pipe 3 in diameter having more diameter then the velocity will be less but the pressure will be more.

That is why the 3 in diameter is having more pressure than 2 in diameter pipe.

Therefore the answer will be a.

a) 3-in diameter pipe

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A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wave

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Answer:

$\lambda$ = 40 cm

Explanation:

given data

string length = 30 cm

solution

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3 years ago

A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im

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Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

$\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}$

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}$

$\dfrac{1}{v}=\dfrac{7}{690}$

$v=98.57\ cm$

We need to calculate the height of the image

Using formula of height

$\dfrac{h'}{h}=\dfrac{-v}{u}$

Where, h' = height of image

h = height of object

$\dfrac{h'}{4}=\dfrac{-98.57}{-30}$

$h'=\dfrac{98.57\times4}{30}$

$h'=13.14\ cm$

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}$

$\dfrac{1}{v}=-\dfrac{10}{299}$

$v=-29.9\ cm$

We need to calculate the height of the image

$\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}$

$h=-\dfrac{29.9\times4}{13.0}$

$h=-9.2\ cm$

The image is virtual and erect.

Hence, This is required solution.

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4 years ago

I. )Which unit can be used to express force?

frutty [35]

The correct choices, sequentially and respectively, are c., d., d., a., d.

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3 years ago

A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging

bekas [8.4K]

Answer:

The answers to the question are

(a) 2.1 m/s

(b) 0.83 N

(c) 1.9 N

Explanation:

To solve the question, we list out the varibles

Length, l of string = 0.8 m

mass of rock, m = 0.12 kg

Angle with the verrticakl, θ = 45 °

a) To find the speed of the rock when the string passes through the vertical position we have

From the first law of thermodynamics

Potential energy = kinetic energy

m×g×l×(1-cosθ) = 1/2×m×v²

That is v² = 2×g×l×(1-cosθ)

= 2×9.81×0.8×(1-cos45) = 4.597

or v = √4.597 = 2.1 m/s

(b) The tension in the string when it makes an angle of 45∘ with the vertical is given by

For balance between Tension and mass of rock is gigen by

∑Forces = 0, T - m×g×cosθ = 0

or T = m×g×cosθ = 0.12×9.81×cos45 = 0.83 N

c) The tension in the string as it passes through the vertical

when passing through the vertical we have T - m×g = (m×v²)/r

or T = m×g + (m×v²)/r = mg(1+2(1-cosθ)) =0.981*0.12 (1+ 2(1-cos45)) =1.867 N

= 1.9 N

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3 years ago