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Anuta_ua [19.1K]
3 years ago
8

Equal masses are suspended from two separate wires made of the same material. The wires have identical lengths. The first wire h

as a larger cross-sectional area than the second wire. Which wire will stretch the LEAST?
a. Both wires will stretch the same amount.
b. The second wire will stretch the least.
c. The first wire will stretch the least.
d. The answer cannot be determined from the information given.
Physics
1 answer:
choli [55]3 years ago
6 0
D. I think is the correct answer
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The water is flowing through the horizontal constricted pipe. The pressure at one end is 4500Pa, speed is 3m/s and area of cross
Oksana_A [137]

The speed and pressure at another end will be 9 m/sec and -31500 pascals.

<h3>What is gauge pressure?</h3>

The difference between absolute pressure and atmospheric pressure is known as gauge pressure. Relative pressure is another name for gauge pressure.

Given data in problem is;

For horizontal pipe, Z₁=Z₂

Pressure at end 1, P₁ = 4500Pa

Speed at end 1, V₁ = 3 m/sec

Speed at end 2, V₂ = ? m/sec

Area of cross-section at end 1,A₁ =A

Area of cross-section at end 1,A₂ = A/3

From the continuity equation;

A₁V₁=A₂V₂

A× 3  = (A/3)×V₂

V₂ = 9 m/sec

From Bernoulli's equation;

\rm \frac{P_1}{\rho g} + \frac{v_1^2}{2g} +Z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} +Z_2  \\\\ \frac{P_1-P_2}{\rho g } = \frac{v_2^2-v_1^2}{2g} \\\\ \frac{P_1-P_2}{\rho} =  \frac{v_2^2-v_1^2}{2} \\\\ \frac{4500-P_2}{1000} =  \frac{9^2-3^2}{2} \\\\ 4500 -P_2 = 36000 \\\\ P_2 =  - \ 31,500 \ pa

Hence the speed and pressure at another end will be 9 m/sec and -31500 pascals.

To learn more about the gauge pressure, refer to the link;

brainly.com/question/14012416

#SPJ1

7 0
2 years ago
Which scenario did not include a chemical change?
Jobisdone [24]

Answer:

what scenario i dont understand

Explanation:

step by step explenation

7 0
3 years ago
A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration
Eduardwww [97]
my answer is 1276.7 because I subtracted 88.3 miles and seconds from 1365 MI
7 0
3 years ago
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 5.0 km/h
Ratling [72]

Answer

given,

time  = 10 s

ship's speed = 5 Km/h

F = m a

a is the acceleration and m is mass.

In the first case

F₁=m x a₁

where a₁ =  difference in velocity / time

F₁ is constant acceleration is also a constant.

Δv₁ = 5 x 0.278

Δv₁ = 1.39 m/s

a_1=\dfrac{1.39}{10}

a₁ = 0.139 m/s²

F₂ =m x a₂

F₃ = F₂ + F₁

Δv₃ = 19 x 0.278

Δv₃ = 5.282 m/s

a₃=Δv₂ / t

a_3=\dfrac{5.282}{10}

a₃ = 0.5282 m²/s

m a₃=m a₁ + m a₂

a₃ = a₂ + a₁

0.5282 = a₂ + 0.139

a₂=0.3892 m²/s

F₂ = m x 0.3892...........(1)

F₁ = m x 0.139...............(2)

F₂/F₁

ratio = \dfrac{0.3892}{0.139}

ratio = 2.8

6 0
4 years ago
An object moves with constant acceleration 3.10 m/s2 and over a time interval reaches a final velocity of 12.4 m/s. If its initi
Harrizon [31]

Answer:

Explanation:

Given:

a = 3.10 m/s^2

vf = 12.4 m/s

vi = -6.2 m/s

t = (vf - vi)/a

= (12.4 + 6.2)/3.1

= 6 s

displacement = (vf - vi)*t

= (12.4 + 6.2) * 6

= 111.6 m.

3 0
3 years ago
Read 2 more answers
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