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3 years ago

6

60 POINTS SO PLEASE HELP ME To add vectors in a straight line and are considered positive directio

ns.
______________ and ______________ are considered negative directions.

1 answer:

tankabanditka [31]3 years ago

7 0

Answer:

The answer to your question is

Explanation:

To add vectors in a straight line _<u>right</u>_________ and ___<u>upwards</u>_______ are considered positive directions. __<u>left</u>________ and __<u>downwards</u>____________ are considered negative directions.

I hope it helps you

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A 1.80-m -long uniform bar that weighs 531 N is suspended in a horizontal position by two vertical wires that are attached to th

Readme [11.4K]

Answer:

(a) 498.4 Hz

(b) 442 Hz

Solution:

As per the question:

Length of the wire, L = 1.80 m

Weight of the bar, W = 531 N

The position of the copper wire from the left to the right hand end, x = 0.40 m

Length of each wire, l = 0.600 m

Radius of the circular cross-section, R = 0.250 mm = $.250\times 10^{- 3}\ m$

Now,

Applying the equilibrium condition at the left end for torque:

$T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}$

$T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}$

$T_{C} = 341.357\ Nm$

The weight of the wire balances the tension in both the wires collectively:

$W = T_{Al} + T_{C}$

$531 = T_{Al} + 341.357$

$T_{Al} = 189.643\ Nm$

Now,

The fundamental frequency is given by:

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

$\mu = A\rho = \pi R^{2}\rho$

(a) For the fundamental frequency of Aluminium:

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}$

where

$\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz$

(b) For the fundamental frequency of Copper:

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}$

where

$\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz$

7 0

3 years ago

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$

Surface area:

$A=P*L=0.04m*1m=0.04m^2$

Then, the heat Q is:

$600\frac{W}{m^2} *0.04m^2=24W$

Finally, find the exit temperature:

$T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077$

$T_{out}$ =27.0000077 ºC

The temperature change so little because:

The mass flow is so big compared to the heat flux.
The transfer area is so little, a bigger length would be required.

3 0

3 years ago

This question looks so simple , but I don’t want to get it won’t could I get help .

Doss [256]

Answer:

I would think a vector but double check that before turning it in

Explanation:

4 0

3 years ago

Read 2 more answers

n general, which trophic level has the LEAST energy available to it? A) producer B) primary consumer C) secondary consumer D) te

zaharov [31]

D - tertiary consumer

This is because it is the farther up to food chain.

4 0

3 years ago

Contrast the force of gravity between these pairs of objects a 1 kg mass and a 2 kg mass that are 1 m apart and two 2 kg masses

Yuki888 [10]

Solution

Force between pair of objects of masses 1kg and 2kg that are 1m apart is given as

$F=G\times \frac{1\times 2}{1^2}$

here G is gravitational constant

G= $6.674\times10^{-11}Nm^2kg^{-2}$

therefore,

$F=6.674\times10^{-11}Nm^2kg^{-2}\times 2$

$F=13.348\times10^{-11}Nm^2kg^{-2}$

similarly Force between pair of objects of masses 2kg each that are 1m apart is given as

$F'=G\times \frac{2\times 2}{1^2}$

$F'=6.674\times10^{-11}Nm^2kg^{-2}\times 4$

$F'=2\times 13.348\times10^{-11}Nm^2kg^{-2}$

or F'=2F

it means Force between pair of objects of masses 2kg each that are 1m apart is equal to twice the Force between pair of objects of masses 1kg and 2kg that are 1m

7 0

3 years ago