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Eddi Din [679]
3 years ago
13

17. The four points built into the camera to ensure that each exposure can be oriented

Physics
1 answer:
3241004551 [841]3 years ago
7 0
The diameter of the circle is 18 m. Eugene incorrectly says that the circumference of the circle is about 113.04 m. What mistake did Eugene make? Use 3.14 for pi.
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Which of the following will attract one another? A. The north pole of magnet and the south pole of another magnet B. The north p
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You raise a bucket of water from the bottom of a deep well. if your power output is 100 w, and the mass of the bucket and the wa
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An oscillating object takes 0.10 s to complete one cycle; that is, its period is 0.10 s. what is its frequency f? express your a
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5 0
3 years ago
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
jekas [21]

Answer:

a)  1.301 kg/s

b) 0.001301 m³/s

c) V₁ = 6.505 m/s, V₂ = 1.626 m/s

d) 118.93 kPa

Explanation:

Given:

The number of cans  = 220

The volume of can, V = 0.355 L = 0.355 × 10⁻³ m³

time = 1 minute = 60 seconds

gauge pressure at point 2, P₂ = 152 kPa

b) Thus, the volume flow rate, Q = Volume/ time

Q = (220 × 0.355 × 10⁻³)/60 = 0.001301 m³/s

a) mass flow rate = Volume flow rate × density

since it is mostly water, thus density of the drink = 1000 kg/m³

thus,

mass flow rate = 0.001301 m³/s × 1000 kg/m³ = 1.301 kg/s

c) Given:

Cross section at point 1 = 2.0 cm² = 2 × 10 ⁻⁴ m²

Cross section at point 2 = 8.0 cm² = 8 × 10 ⁻⁴ m²

also,

Q = Area × Velocity

thus, for point 1

0.001301 m³/s = 2 × 10 ⁻⁴ m² × velocity at point 1 (V₁)

or

V₁ = 6.505 m/s

for point 2

0.001301 m³/s = 8 × 10 ⁻⁴ m² × velocity at point 1 (V₂)

or

V₂ = 1.626 m/s

d) Applying the Bernoulli's theorem between the points 1 and 2 we have

P_1+\rho gV_1 + \frac{\rho V_1^2}{2}=P_2+\rho gV_2 + \frac{\rho V_2^2}{2}

or

P_1=P_2+\rho\timesg(y_2-y_1)+\frac{\rho}{2}(V_2^2-V_1^2))

on substituting the values in the above equation, we get

P_1=152+1000\times 9.8(1.35)+\frac{1000}{2}(1.626^2-6.505^2))

it is given that point 1 is above point 2 thus, y₂ -y₁ is negative

or

P_1=118.93\ kPa

thus, gauge pressure at point 1 is 118.93 kPa

8 0
3 years ago
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