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3 years ago

Flat plate collector can provide temperature upto_____

Engineering

2 answers:

Vinil7 [7]3 years ago

5 0

Answer:

Explanation:

<h3>Flat plate collector can provide temperature upto 80°C </h3>

<h3>Hope my answer is helpful to you ❣️☪️❇️❇️☪️❣️✌️</h3>

lana66690 [7]3 years ago

4 0

Answer:

80⁰C

Explanation:

80°C.

Normal flat plate collectors can deliver heat at temperatures up to 80°C. Deficiency rates for normal flat plate collectors can be classified as visual losses, which produce with cumulative angles of the incident sunshine, and thermal losses, which upsurge fast with the working temperature intensities

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8. Two 40 ft long wires made of differing materials are supported from the ceiling of a testing laboratory. Wire (1) is made of

san4es73 [151]

Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

Material H has a modulus of elasticity E=1.009 × 10⁹ Pa

Material K has higher value of modulus of elasticity than material H

Material K is stiffer.

Explanation:

Wire 1 material H

Length=L = 40 ft =12.192 m

Diameter= 3/8 in = 0.009525 m

Area= A= πr²,where r=0.009525/2 =0.004763

A=3.142*0.004763² =0.00007126 m²

Force, F= 225 lb= 225*4.45 =1001.25 N

Change in length =Δ L= 0.10 in = 0.00254

To find modulus of elasticity apply'

E=F*L/A*ΔL

E=1001.25*12.192/(0.004763*0.00254)

E= 1009027923.58 Pa

E=1.009 × 10⁹ Pa

For Wire 2 material K

Length=L= 40 ft =12.192 m

Diameter = 3/16 in = 0.1875 in = 0.004763 m

Area= πr² = 3.142 * (0.004763/2)² = 0.00000567154 m²

Force, F= 225 lb= 225*4.45 =1001.25 N

Change in length =Δ L= 0.25 in =0.00635 m

To find modulus of elasticity apply'

E=F*L/A*ΔL

E= (1001.25*12.192)/(0.00000567154 * 0.00635 )

E=338955422575 Pa

E=3.389× 10¹¹ Pa

Material K has a greater modulus of elasticity

The material with higher value of E is stiffer than that with low value of E.The stiffer material is K.

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3 years ago

A group of n Ghostbusters is battling n ghosts. Each Ghostbuster carries a proton pack, which shoots a stream at a ghost, eradic

babunello [35]

Answer:

Using the above algorithm matches one pair of Ghostbuster and Ghost. On each side of the line formed by the pairing, the number of Ghostbusters and Ghosts are the same, so use the algorithm recursively on each side of the line to find pairings. The worst case is when, after each iteration, one side of the line contains no Ghostbusters or Ghosts. Then, we need n/2 total iterations to find pairings, giving us an P( $n^{2} lg n$ )- time algorithm.

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4 years ago

The capacity of a battery is 1800 mAh and its OCV is 3.9 V. a) Two batteries are placed in series. What is the combined battery

Lynna [10]

Answer:

capacity = 0.555 mAh

capacity = 3600 mAh

Explanation:

given data

battery = 1800 mAh

OCV = 3.9 V

solution

we get here capacity when it is in series

so here Q = 2C

capacity = 2 × ampere × second ...............1

put here value and we get

and 1 Ah = 3600 C

capacity = $\frac{2}{3600}$

capacity = 0.555 mAh

and

when it is in parallel than capacity will be

capacity = Q1 +Q2 ...............2

capacity = 1800 + 1800

capacity = 3600 mAh

3 0

4 years ago

An 1,840 W toaster, a 1,420 W electric frying pan, and a 70 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (T

Viktor [21]

Answer:

Current drawn by toaster = 15.33 A

Current drawn by electric frying pan = 11.83 A

Current drawn by lamp = 0.58 A

The fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.

Explanation:

There are three devices plugged into the same outlet.

Toaster = 1840 W

Electric frying pan = 1420 W

Lamp = 70 W

Since the three devices are connected in parallel therefore, the voltage across them will be same but the current drawn by each will be different.

A) What current is drawn by each device?

The current flowing through the device is given by

I = P/V

Where P is the power and V is the voltage.

Current drawn by toaster:

I = 1840/120

I = 15.33 A

Current drawn by electric frying pan:

I = 1420/120

I = 11.83 A

Current drawn by lamp:

I = 70/120

I = 0.58 A

B) Will this combination blow the 15-A fuse?

The total current drawn by all three devices is

Total current = 15.33 + 11.83 + 0.58

Total current = 27.74 A

Therefore, the fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.

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4 years ago

A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima

zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

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