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3 years ago

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Suppose that units of force, length, and time are chosen such that the density of water and the acceleration of gravity are both

of unit magnitude. If the pound is taken as the unit of force, what are the sizes of the units of length and time?

1 answer:

k0ka [10]3 years ago

3 0

Answer: So a 1 pound = 0.453592kg

the density of the water is D= 997 kg/ $m^{3}$ so we need to use replace the kg with pounds with 2.20462 pounds = 1kg.

then D = 2.20462*997 pounds/ $m^{3}$

and we want to make this equal to one changing the metters for other thing.

we need to replace $x^{3}$ with 2.020462*997 meters qube.

then our new metters variable x = 12.6meters.

now g = 9.8 meters for second square.

in our new lenght variable x.

g = 12.6*9.8 x for second square.

so if y is our new time variable

$y^{2}$ = 12.6*9.8 seconds square. so y = 11.1seconds

so the units of length are x = 12.6 meters and for time y = 11.1 seconds.

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A sinusoidal wave of frequency 420 Hz has a speed of 310 m/s. (a) How far apart are two points that differ in phase by π/8 rad?

Olin [163]

Answer:

a) Two points that differ in phase by π/8 rad are 0.0461 m apart.

b) The phase difference between two displacements at a certain point at times 1.6 ms apart is 4π/3.

Explanation:

f = 420 Hz, v = 310 m/s, λ = wavelength = ?

v = fλ

λ = v/f = 310/420 = 0.738 m

T = periodic time of the wave = 1/420 = 0.00238 s = 0.0024 s = 2.4 ms

a) Two points that differ in phase by π/8 rad

In terms of the wavelength of the wave, this is equivalent to [(π/8)/2π] fraction of a wavelength,

[(π/8)/2π] = 1/16 of a wavelength = (1/16) × 0.738 = 0.0461 m

b) two displacements at times 1.6 ms apart.

In terms of periodic time, 1.6ms is (1.6/2.4) fraction of the periodic time.

1.6/2.4 = 2/3.

This means those two points are 2/3 fraction of a periodic time away from each other.

1 complete wave = 2π rad

Points 2/3 fraction of a wave from each other will have a phase difference of 2/3 × 2π = 4π/3.

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4 years ago

hree large plates are separated bythin layers of ethylene glycol and water. The top plate moves to the right at 2m/s. At what sp

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Answer: For the center plate to remain stationed in one position without rotating, the bottom plate has to move to the left at a speed of 2m/s, so as to cancel the force acting on it from the top.

The center plate will not move when the bottom plate is moving left in a speed of 2m/s to counter the speed of the top plate, because a body will continue to be at rest if all the forces acting towards the body are equal. The center plate will be at rest because we have directed equal force from the top and bottom of the plate.

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3 years ago

A dryer is shaped like a long semi-cylindrical duct of diameter 1.5 m. The base of the dryer is occupied with water-soaked mater

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Answer:

0.0371 kg/s.m

Explanation:

From the given information, let's have an imaginative view of the semi-cylinder; (The image is shown below)

Assuming the base surface of both ends of the cylinder is denoted by:

$A_1 \ and \ A_2$

Thus, using the summation rule, the view factor $F_{11$ and $F_{12$ is as follows:

$F_{11}+F_{12}=1$

Let assume the surface (1) is flat, the $F_{11} = 0$

Now:

$0+F_{12}=1$

$F_{12}=1$

However, using the reciprocity rule to determine the view factor from the dome-shaped cylinder $A_2$ to the flat base surface $A_1$ ; we have:

$A_2F_{21} = A_{1}F_{12} \\ \\ F_{21} = \dfrac{A_1}{A_2}F_{12}$

Suppose, we replace DL for $A_1$ and

$A_2$ = $\dfrac{\pi D}{2}$

Then:

$F_{21} = \dfrac{DL}{(\dfrac{\pi D}{2}) L} \times 1 \\ \\ =\dfrac{2}{\pi} \\ \\ =0.64$

Now, we need to employ the use of energy balance formula to the dryer.

i.e.

$Q_{21} = Q_{evaporation}$

But, before that; let's find the radian heat exchange occurring among the dome and the flat base surface:

$Q_{21}= F_{21} A_2 \sigma (T_2^4-T_1^4) \\ \\ Q_{21} = F_{21} \times \dfrac{\pi D}{2} \sigma (T_2^4 -T_1^4)$

where;

$\sigma = Stefan \ Boltzmann's \ constant$

$T_1 = base \ temperature$

$T_2 = temperature \ of \ the \ dome$

∴

$Q_{21} = 0.64 \times (\dfrac{\pi}{2}\times 1.5) \times 5.67 \times 10^4 \times (1000^4 -370^4)\\ \\ Q_{21} = 83899.15 \ W/m$

Recall the energy balance formula;

$Q_{21} = Q_{evaporation}$

where;

$Q_{evaporation} = mh_{fg}$

here;

$h_{fg}$ = enthalpy of vaporization

m = the water mass flow rate

∴

$83899.15 = m \times 2257 \times 10^3 \\ \\ m = \dfrac{83899.15}{ 2257 \times 10^3 }\\ \\ \mathbf{m = 0.0371 \ kg/s.m}$

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A liquid propellant engine has the following characteristics: chamber pressure of 7 MPa, constant ratio of specific heats of 1.3

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Answer:

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1) Thrust coefficient at sea level.

Cfsl = TSL / Pca

TSL = Mp * Vc + ( Pc - Pa )Ac

Mp = mass flux = 43.75 kg/s

∴ Cfsl = Mp Vc / Pca + ( Pc - Pa )/Pc * ( Ac / A* )

= 1.6 - 0.04923 = 1.55

<u>2) Specific impulse at sea </u>

Isp = Vc / g = 2549.75 / 9.81

= 260 N.s

3) Altitude at optimal expansion

H = 3370 m

<u>4) thrust coefficient at optimal expansion </u>

CF = 1.6

attached below is the detailed solution

<u>5) Mass flux through the throat </u>

Mass flux = P1 * At / Cc

= ( 7*10^6 * 0.01 ) / 1600

= 43.75 kg/s

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Primary coil

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