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2 years ago

10

A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva

so you can better study its flow pattern in the laboratory. Your model must be able to move at 50 cm/s and you will place the model in honey instead of water. Honey has a density of 1400 kg/m3 and a viscosity of 600 Pa-s.
Required:
How long should your model be?

1 answer:

AleksandrR [38]2 years ago

5 0

Answer:

Explanation:

For the problem, we should have same reynolds number

ρvd/mu = constant

1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600

d = 25.66 cm

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Calculate the mass of the air contained in a room that measures 2.50 m x 5.50 m x 3.00 m if the density of air is 1.29 g/dm3.53.

Law Incorporation [45]

Answer:

$5.32\cdot 10^4 g$

Explanation:

First of all, we need to find the volume of the room, which is given by

$V=2.50 m \cdot 5.50 m \cdot 3.00 m =41.3 m^3$

Now we can find the mass of the air by using

$m=dV$

where

$d=1.29 g/dm^3$ is the density of the air

$V=41.3 m^3 = 41,300 dm^3$ is the volume of the room

Substituting,

$m=(1.29)(41300)=5.32\cdot 10^4 g$

6 0

2 years ago

The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2

masha68 [24]

Answer:

the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

Explanation:

Given that:

Modulus for elasticity for concrete post $E_c$ = $3.6 *10^6$ psi

Thermal coefficient for concrete post $\alpha _c$ = $5.5 *10^{-6}/^0F$

Modulus for elasticity of steel bar $E_s$ = $29*10^6psi$

Thermal coefficient of steel bar $\alpha _2$ = $6.5*10^{-6}/^0F$

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod $A_s$ = $6(\frac{ \pi}{4} )(d_s)^2$

= $6(\frac{ \pi}{4} )(\frac{7}{8} )^2$

= 3.61 in²

Area of concrete parts $A_c$ = (10)(10) - $A_s$

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= $[\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)$

= $[\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)$

= $[(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}$

= $1.234*10^{-8}P = 8.0*10^{-5}$

$P = \frac {8.0*10^{-5}}{1.234*10^{-8}}$

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

$\sigma_c =\frac{P}{A_c}$

$\sigma_c = \frac{6482.98}{96.39}$

$\sigma_c = 67.26 psi$

Thus, the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

Also; the normal stress in the steel bars induced as a result of temperature rise is as follows:

$\sigma_s = \frac{-P}{A_s}$

$\sigma_s =\frac{-6482.98}{3.61}$

$\sigma_s = - 1795.84 psi$

Thus, the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

6 0

2 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

2 years ago

Jeff of the Jungle swings on a 7.6-m vine that initially makes an angle of 32 ∘ with the vertical.

shtirl [24]

To solve this problem we will use the trigonometric concepts to find the distance h, which will allow us to find the speed of Jeff and that will finally be the variable that will indicate the total tension, since it is the variable of the centrifugal Force given in the vine at the lowest poing of the swing.

From the image:

$cos (32) = \frac{(7.6-h)}{7.6}$

$h = 1.1548m$

When Jeff reaches his lowest point his potential energy is converted to kinetic energy

$PE = KE$

$mgh = \frac{1}{2} mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8)(1.1548)}$

$v= 4.75m/s$

Tension in the string at the lowest point is sum of weight of Jeff and the his centripetal force

$T = W+F_c$

$T = mg + \frac{mv^2}{r}$

$T = (83)(9.8)+\frac{(9.8)( 4.75)^2}{7.6}$

$T = 842.49N$

Therefore the tension in the vine at the lowest point of the swing is 842.49N

3 0

2 years ago

Marcus can drive his boat 24 miles down the river in 2 hours but takes 3 hours to return upstream. Find the rate of the boat in

notsponge [240]

Answer:

speed of boat as

$v_b = 10 mph$

river speed is given as

$v_r = 2 mph$

Explanation:

When boat is moving down stream then in that case net resultant speed of the boat is given as

since the boat and river is in same direction so we will have

$v_1 = v_r + v_b$

Now when boat moves upstream then in that case the net speed of the boat is opposite to the speed of the river

so here we have

$v_2 = v_b - v_r$

as we know when boat is in downstream then in that case it covers 24 miles in 2 hours

$v_1 = \frac{24}{2} = 12 mph$

also when it moves in upstream then it covers same distance in 3 hours of time

$v_2 = \frac{24}{3} = 8 mph$

$v_b + v_r = 12 mph$

$v_b - v_r = 8 mph$

so we have speed of boat as

$v_b = 10 mph$

river speed is given as

$v_r = 2 mph$

8 0

2 years ago

Read 2 more answers