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3 years ago

1.33 m^3 of fluid flows out of a pipe in 24.5 s. the fluid leaves the pipe at 3.55 m/s. what is the area of the pipe. (unit=m^2)

Physics

2 answers:

denpristay [2]3 years ago

8 0

Correct Answer:

$0.0154 m^2$

Margaret [11]3 years ago

3 0

Answer:

0.015 $meter^{2}$

Explanation:

Total volume of water coming out = 1.33 $meter^{3}$

Also volume = Cross sectional area*Length covered

Length covered = Velocity *time

=24.5*3.55

=86.97 meter

Let the cross sectional area be A.

1.33 = 86.97*A

A =0.015 $meter^{2}$

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Rony fills a bucket with water and whirls it in a vertical circle to demonstrate that the water will not spill from the bucket a

Romashka-Z-Leto [24]

Answer:

0 N, 3.49 m/s

Explanation:

Draw a free body diagram for the bucket at the top of the swing. There are two forces acting on the bucket: weight and tension, both downwards.

If we take the sum of the forces in the radial direction, where towards the center is positive:

∑F = ma

W + T = m v² / r

The higher the velocity that Rony swings the bucket, the more tension there will be. The slowest he can swing it is when the tension is 0.

W = m v² / r

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 1.24 m:

v = √(9.8 m/s² × 1.24 m)

v = 3.49 m/s

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3 years ago

Which statement is true about tomato juice? A) It is an acid. B) It has a pH of 7.0. Eliminate C) It is strongly basic in nature

adoni [48]

The answer would be A. It is an acid.

7 0

3 years ago

A roller coaster is stationary at the top of a ramp.

Serjik [45]

Answer:

Explanation:

The roller coaster is stationary so the kinetic energy would be zero, but it is at the top of ramp so the potential energy would be high as its gravitational so it would have to be A

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3 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

Explain how you should walk

hammer [34]

Answer:

A step by step to walk

Explanation:

One- Make sure your shoes are tied so that you dont trip

Two- Make sure your way is a cleared path so you dont fall or even hurt yourself

three- use both set of lets to go in any direction you want.

Four- when walking make sure to try and keep a steedy pace so that both set of legs are going up and down but in harmony

3 0

3 years ago

Read 2 more answers