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3 years ago

Drag each label to the correct location on the image.

Engineering

2 answers:

Thepotemich [5.8K]3 years ago

7 0

Answer:

The correct classification of characteristics according to the camera model is:

Professional Camera:

has interchangeable lenses

uses charge-coupled devices (CCD)

provides quality videos at a great price is bigger and heavier and may need additional support to carry

Consumer Camera:

provides quality videos at a great price

is mostly automatic

records using tape only

Explanation:

The reasons behind this answer are the following but first of all, we need to say that we are talking about very old cameras. Around 20-year-old gear, so this is very inaccurate to modern cameras. Now, the first reason to classify these matches like we have done it is that in the first place professional cameras that used CCD were really big and required additional supports to be carried, such as tripods, chest munt, or even little cars called dollies. Now we use scorpions and something smaller and lighter. Second, consumer cameras only recorded in tape, back in the day we had no DVD, no memory stick, or even hard drive. The only tape for consumer cameras as the other resources were too expensive to be included on them.

Misha Larkins [42]3 years ago

6 0

The guy above got me this question wrong

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A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a

Dafna1 [17]

Answer:

a) $\dot W = 1.062\,kW$

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}$

$COP_{HP} = 15.692$

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

$\dot Q_{H} = 60000\,\frac{kJ}{h}$

The minimum power supplied to the heat pump is:

$\dot W = \frac{\dot Q_{H}}{COP}$

$\dot W = \frac{\left(60000\,\frac{kJ}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)}{15.692}$

$\dot W = 1.062\,kW$

5 0

3 years ago

If something is 50fficient, how many joules of wasted energy will there be if 750j of energy is put in?’

Sedbober [7]

Efficiency is the minimum use of energy to accomplish the task. The wasted energy will be 375 J when 750 J of energy is given.

<h3>What is wasted energy?</h3>

Wasted energy is energy that is not useful when the transformation in the system occurs.

Total energy = 750 J

The efficiency of the system = 50 %

Output work (OW) is calculated as:

Efficiency = output work ÷ input work × 100%

750 × 50 = 100 OW

OW = 375 J

Wasted energy = Total energy - output work

= 750 - 375

= 375 J

Therefore, the machine is 50 % inefficient and has wasted energy of 375 J.

Learn more about wasted energy here:

brainly.com/question/16177264

#SPJ4

8 0

2 years ago

A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir

Olenka [21]

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

= 500 - 2(20)

= 460 mm

So, internal radius, r = 230 mm

= 0.23 m

Density of molten metal, ρ = $7.2 \ g/cm^3$

= $7200 \ kg/m^3$

The height of pouring cavity above parting surface is h = 300 mm

= 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

Area, A $=2 \pi r^2$

$=2 \pi (0.23)^2$

$=0.3324 \ m^2$

$F=\rho g A h$

$=7200 \times 9.81 \times 0.3324 \times 0.3$

= 7043.42 N

3 0

3 years ago

Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E

White raven [17]

Answer:

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

Explanation:

To solve this problem we use the expression for the temperature film

$T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5$

Then, we have to compute the Reynolds number

$Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}$

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

$Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77$

but we also now that

$Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\$

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0

4 years ago

The amplitudes of the displacement and acceleration of an unbalanced motor were measured to be 0.15 mm and 0.6*g, respectively.

ehidna [41]

Answer:

The speed of shaft is 1891.62 RPM.

Explanation:

given that

Amplitude A= 0.15 mm

Acceleration = 0.6 g

we can say that acceleration= 0.6 x 9.81

$acceleration,a=5.88\ \frac{m}{s^2}$

We know that

$a=\omega ^2A$

So now by putting the values

$a=\omega ^2A$

$5.88=\omega ^2 \0.15\times 10^{-3}$

$\omega =198.09\ \frac{rad}{s}$

We know that

ω= 2πN/60

198.0=2πN/60

N=1891.62 RPM

So the speed of shaft is 1891.62 RPM.

4 0

3 years ago