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Brilliant_brown [7]
3 years ago
14

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per min

ute. At point 2 in the pipe, the gauge pressure is 152kPa and the cross-sectional area is 8.00cm2. At point 1, 1.35m above point 2, the cross-sectional area is 2.00cm2.
a. Find the mass flow rate.
b. Find the volume flow rate.
c. Find the flow speed at point 1.
d. Find the flow speed at point 2.
e. Find the gauge pressure at point
Physics
1 answer:
Yuki888 [10]3 years ago
7 0

Answer:

A. To find the mass flow rate.

We use= 220 x 0.355/ 60

= 1.3kg/s

B. Volume flowrate is = mass flowrate / density

But density is 1000kg/m³

= 1.3kg/s/ 1000kg/m³

= 0.0013m³/s

C. Flow speead at 1

= 0.0013m³/s / (2 x 10-2m)²

= 6.5m/s

D.flow speed at 2

0.0013m³/s / (8x 10-2m)²

=1.63m/s

E. Gauge pressure at point 1

= 152+ 1/1000 ( 1.63)²- 6.5² + 1000( 9.8) ( 0-1.35)

= 119kpa

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The critical angle for a certain air-liquid surface is 47.7°. What is the index of refraction of the liquid? Round to the neares
KengaRu [80]

Answer:

The index of refraction of the liquid is 1.35.

Explanation:

It is given that,

Critical angle for a certain air-liquid surface, \theta_1=47.7^{\circ}

Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1

Using Snell's law for air liquid interface as :

n_1\ sin\theta_1=n_2\ sin(90)

n_1\ sin(47.7)=1

n_1=\dfrac{1}{sin(47.7)}

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3 years ago
How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un
Keith_Richards [23]

Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is 4.20\times10^{8}\ kg-m/s.

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

W=K.E

W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2

Put the value into the formula

W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2

W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2

W=1.122\times10^{-9}\ J

W=7001\ MeV

(b). We need to calculate  the momentum of this proton

Using formula of momentum

p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}

Put the value into the formula

p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}

p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}

p=1.404\times10^{-26}c

p=4.20\times10^{8}\ kg-m/s

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is 4.20\times10^{8}\ kg-m/s.

4 0
3 years ago
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