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3 years ago

Freeee Poinntssss 100!!!!!! Hi how you doin?

Engineering

2 answers:

Andreas93 [3]3 years ago

6 0

Thank u very much , im doin good wby :)

Zinaida [17]3 years ago

3 0

Answer:

Explanation:

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Describe at least one way you take advantage or could take advantage of each of the different forms of energy

Scorpion4ik [409]

Answer:

Kinetic energy can be used to develop electric energy which can be used as electricity.

Explanation:

The kinetic energy can be harnessed; much like some hydro power technologies harness water movement. A way to convert this kinetic energy into electric energy is through piezoelectric. By applying a mechanical stress to a piezoelectric crystal or material an electric current will be created and can be harvested.

Kinetic energy is also generated by the human body when it is in motion. Studies have also been done using kinetic energy and then converting it to other types of energy, which is then used to power everything from flashlights to radios and more.

4 0

3 years ago

Two units for measuring magnetic flux density are:

romanna [79]

The correct answer is A
Faraday and Weber
:)

5 0

2 years ago

A cubical picnic chest of length 0.5 m, constructed of sheet styrofoam of thickness 0.025 m, contains ice at 0\[Degree]C. The th

maksim [4K]

Answer:

Rate of heat transfer is 0.56592 kg/hour

Explanation:

Q = kA(T2 - T1)/t

Q is rate of heat transfer in Watts or Joules per second

k is thermal conductivity of the styrofoam = 0.035 W/(mK)

A is area of the cubical picnic chest = 6L^2 = 6(0.5)^2 = 6×0.25 = 1.5 m^2

T1 is initial temperature of ice = 0 °C = 0+273 = 273 K

T2 is temperature of the styrofoam = 25 °C = 25+273 = 298 K

t is thickness of styrofoam = 0.025 m

Q = 0.035×1.5(298-273)/0.025 = 1.3125/0.025 = 52.5 W = 52.5 J/s

Mass flow rate = rate of heat transfer ÷ latent heat of melting of ice = 52.5 J/s ÷ 3.34×10^ 5 J/kg = 1.572×10^-4 kg/s = 1.572×10^-4 kg/s × 3600 s/1 hr = 0.56592 kg/hr

6 0

3 years ago

Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t

alukav5142 [94]

Answer:

hoop stress
longitudinal stress
material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = $\frac{PD}{2T}$ EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = $\frac{PD}{4T}$ EQUATION 2

where p = water pressure inside the hose

d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

$\frac{D}{T}$ ≥ 20

insert this value into equation 1

αh = $\frac{20 *30}{2}$ = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

8 0

3 years ago

The temperature at the bottom of a reservoir is TL = 280 K and the surface temperature is TH = 295 K. This temperature differenc

Tanzania [10]

a) For the thermal efficiency we have

$\eta_{th} = \frac{Q_{out}}{Q_{in}} = \frac{|W|}{|Q_h|}\\\eta_{th} = \frac{|W|}{|W|+|Q_2|}$

With the previously values we know that

$W=8kW$ and $Q_L = 1440/6kW$ (convert the min to sec)

Replacing the values

$\eta_{th}=\frac{8}{8+1440/6}=\frac{1}{31}\\\eta_{th}\% = 3.225\%$

b) We use the formula of carnot efficiency

$\eta_{th}=1-\frac{T_l}{T_h}\\\eta_{th}\% =(1-\frac{280}{295})*100\\\eta_{th}\%=5.085\%$

**Note that apply the formula of carnot cycle we need to consider that there is no exchange of heat, there is no friction and the reservior are completely insulated

8 0

3 years ago