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Marizza181 [45]

2 years ago

14

Henry can lift a 200 N load 20 m up a ladder in 40 s. Ricardo can lift twice the load up one-half the distance in the same amoun

t of time. Which comparison between Ricardo and Henry is correct?

2 answers:

prohojiy [21]2 years ago

4 0

Henry will lift 200 N load 20 m up a ladder in 40 s. While the Ricardo will take 400 N load in 80 seconds. So, For Henry to take 400 N load it will take him 80 seconds in two attempts. And,also, he will have to cover 40 m of distance.

madreJ [45]2 years ago

4 0

Answer;

-Ricardo exerted a greater force, did the same amount of work, and had the same power output as Henry.

Explanation;

For Ricardo to lift twice the load lifted by Henry for the one-half the distance in the same amount of time, he will need more force as the load is heavier.

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100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak

Amanda [17]

Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

$Q_m=AC_o\sqrt{2\rho g_cP_g}$

Here, $Q_m$ is the mass flow rate through the leak of the pipeline. $A$ is the area of the hole, $C_o$ is the discharge rate, $\rho$ is the fluid density, $g_c$ is the gravitational constant and $P_g$ is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

$d=(0.1 in)(\frac{1ft}{12in})\\=8.33\times 10^{-3}ft$

Calculate the area (A) of the hole. The area of the hole is.

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

Substitute 3.14 for $\pi$ and $8.33\times 10^{-3}ft$ for d and calculate A.

$A=\frac{\pi d^2}{4}\\\\\frac{(3.14)(8.33\times 10^{-3})^2}{4}\\\\5.45\times 10^{-5}ft^2$

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is $62.4\frac{Ib_m}{ft^3}$ . Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

$=(0.8794)(62.4\frac{Ib_m}{ft^3})\\\\54.9\frac{Ib_m}{ft^3}$

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate $P_g$ .

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

$=\frac{(50psig\times 43ft)+(40psig \times 57ft)}{100ft}\\\\=44.3psig$

Convert the pressure from psig to $Ib_f/ft^2$

$P_g=(44.3psig)(\frac{1\frac{Ib_f}{ft^2}}{1psig})(144\frac{in^2}{ft^2})\\\\=6,379.2\frac{Ib_f}{ft^2}$

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute $5.45\times 10^{-5}ft^2$ for A. 0.61 for $C_o$ , $54.9\frac{Ib_m}{ft^3}$ for $\rho$ , $32.17\frac{ft.Ib_m}{Ib_f.s^2}$ for $g_c$ , and $6,379.2\frac{Ib_f}{ft^2}$ for $P_g$ and calculate $Q_m$

$Q_m=AC_o\sqrt{2\rho g_cP_g}\\\\=(5.45\times 10^{-5}ft^2)(0.61)\sqrt{2(54.9\frac{Ib_m}{ft^3})(32.17\frac{ft.Ib_m}{Ib_f.s^2})(6,379.2\frac{Ib_f}{ft^2})}\\\\(3.3245\times 10^{-5}ft^2)\sqrt{22,533,031.21\frac{Ib^2_m}{ft^4.s^2}}\\\\=0.158\frac{Ib_m}{s}$

The mass flow rate of benzene through the leak in the pipeline is $0.158\frac{Ib_m}{s}$

8 0

2 years ago

Two students grab a slinky and start waving it up and down. A third student counts the number of waves that pass by every second

deff fn [24]

Velocity = frequency * wavelength

v = fλ, Just pick any points on the graph for frequency f and corresponding λ. Taking the first red point at the top. λ = 6m, f = 1 Hz, v = 6 * 1, v = 6 m/s

V = 6 M/S

4 0

2 years ago

Read 2 more answers

A bullet starting from rest accelerates uniformly at a rate of 1,250 meters per square second. What is the bullet's speed after

Tatiana [17]

Answer:

125,000

Explanation:

8 0

2 years ago

An inline skater skates on a circular track 120.0 m in diameter at a tangential speed of 9.20 m/s. If the skater’s mass is 68.5

jok3333 [9.3K]

Answer:

The centripetal force acting on the skater is <u>48.32 N.</u>

Explanation:

Given:

Radius of circular track is, $R=120.0\ m$

Tangential speed of the skater is, $v=9.20\ m/s$

Mass of the skater is, $m=68.5\ kg$

We are asked to find the centripetal force acting on the skater.

We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.

Centripetal force acting on the skater is given as:

$F_c=\frac{mv^2}{R}$

Now, plug in the given values of the known quantities and solve for centripetal force, $F_c$ . This gives,

$F_c=\frac{68.5\times (9.20)^2}{120.0}\\\\F_c=\frac{68.5\times 84.64}{120}\\\\F_c=\frac{5797.84}{120}\\\\F_c=48.32\ N$

Therefore, the centripetal force acting on the skater is 48.32 N.

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2 years ago

A bird flies at an average velocity of 3.60 m/s for 18.4 s. How far does it travel? (unit=m)

jarptica [38.1K]

First, you have to write what you know:

V = 3.60 m/s
D = ?
T = 18.4 s

Next, you plug everything into this formula:

V = D/T
3.60 m/s = ?/18.4 s

Then, you multiply 3.60 m/s and 18.4 s

D = 66.24 m

5 0

2 years ago