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IrinaK [193]
3 years ago
7

Around the turn of the twentieth century, inventor Robert Shuman was engineering products that used energy directly from the Sun

to power motors and farms. He stated that coal and oil would eventually run out and that humans needed to learn how to use the more-abundant energy resource of the Sun. During which stage of technological design did Shuman identify that using oil and coal would be a problem? first second third fourth
Physics
1 answer:
MrRa [10]3 years ago
5 0

Answer: First stage

Explanation:

The four stages of technological design include:

• identifying a problem.

• Designing a solution.

• Implement, build, and test the design.

• Determine if solution met the need.

In this scenario, Robert Shuman identified the problem. This was depicted as he stated that coal and oil would eventually run out and that humans needed to learn how to use the more-abundant energy resource of the Sun,

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34.6 cL= (blank) hL convert
shepuryov [24]

<u>Answer</u>

0.00346 hL

<u>Explanation</u>

cL means Centilitre while hL means Hectolitre.

10,000 cL = 1 hL

∴ 34.6 cL = 34.6/10,000  hL

                = <em>0.00346 hL</em>

3 0
4 years ago
If the magnitude of the electric field in air exceeds roughly 3 ✕ 106 N/C, the air breaks down and a spark forms. For a two-disk
Westkost [7]

Answer:

1.843 x 10^-5 C  

Explanation:

<u><em>Givens:   </em></u>

It is given that the air starts ionizing when the electric field in the air exceeds a magnitude of 3 x 10^6 N/C, which means that the max electric field can stand without forming a spark is 3 x 10^6 N/C.  

Also it is given that the radius of the disk is 50 cm, it is required to find out the max amount of charge that the disk can hold without forming spark, which means the charge that would produce the max magnitude of the electric field that air can stand without forming spark, and since we know that the electric field in between 2 disk "Capacitor" is given by the following equation  

E = (Q/A)/∈o                                (1)

Where,

Q: total charge on the disk.

A: the area of the disk.  

<u><em>Calculations:  </em></u>

We want to find the quantity of charge on the disk that would produce an electric field of 3 x 10^6 N/C, knowing the radius of the disk we can find the cross-section of the disk, thus substituting in equation (1) we find the maximum quantity of charge the disk can hold  

Q = EA∈o

   = (3 x 10^6) x (π*0.50) x (8.85 x 10^-12)  

  = 1.843 x 10^-5 C  

note:

calculations maybe wrong but method is correct

8 0
3 years ago
When atoms are split, they release energy. This concept applies to (2 points)
cupoosta [38]

This applies to nuclear reactions, specifically nuclear fission.

This huge release of energy has been used in atomic bombs and in the nuclear reactors that generate electricity.

╦────────────────────────────╦

│Hope this helped  _____________________│    

│~Xxxtentaction _______________________│

╩__________________________________╩          

6 0
3 years ago
As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury baromet
Pani-rosa [81]

Answer:

new atmospheric pressure is 0.9838 × 10^{5}  Pa

Explanation:

given data

height = 21.6 mm = 0.0216 m

Normal atmospheric pressure = 1.013 ✕ 10^5 Pa

density of mercury = 13.6 g/cm³

to find out

atmospheric pressure

solution

we find first height of mercury when normal pressure that is

pressure p = ρ×g×h

put here value

1.013 × 10^{5}  = 13.6 × 10³ × 9.81 × h

h =  0.759 m

so change in height Δh = 0.759 - 0.0216

new height H = 0.7374 m

so new pressure = ρ×g×H

put here value

new pressure = 13.6 × 10³ × 9.81 × 0.7374

atmospheric pressure = 98380.9584

so new atmospheric pressure is 0.9838 × 10^{5}  Pa

6 0
3 years ago
Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

7 0
3 years ago
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