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Vitek1552 [10]
3 years ago
12

What kind of system is generally composed of a fluid system to move the heat from the collector to its point of usage and a rese

rvoir to stock the heat for future use?
A.
coal burning
B.
nuclear power
C.
solar heating
Physics
1 answer:
Mkey [24]3 years ago
5 0

Solar heating is the system composed of a fluid system to move the heat from the collector to its point of usage and a reservoir to stock the heat

<u>Explanation:</u>

The options given here like coal burning uses solid material as the source to heat and to generate energy. Similarly, nuclear power also requires solid particles like atoms or neutrons to strike the moderators forming energy.

In both of these cases, fluid system is present but it is used completely as coolant and to maintain the temperature. Thus, the remaining system that is solar heating has been done for water tanks where the fluid as water is used to move the heat from its collector to its point of usage. Even in solar system it is used as reservoir to stock the heat.

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Answer:

Explanation:

2 km a hour

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Explanation:

8 0
3 years ago
II Force on a tennis ball. The record speed for a tennis ball that is served is 73.14 m/s. During a serve, the ball typically st
AveGali [126]

Answer:

F=248.5W N

Explanation:

Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

a=\frac{\Delta x}{\Delta t}=\frac{v-v_0}{t-t_0}

Since we start counting at 0s and the ball departs from rest, this is just a=\frac{v}{t}

So we can write:

F=ma=\frac{mv}{t}=\frac{gmv}{gt}

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

F=\frac{Wv}{gt}=\frac{v}{gt}W

Substituting our values:

F=\frac{(73.14m/s)}{(9.81m/s^2)(30\times10^{-3}s)}W=248.5W N

Where the average force exerted has been written it terms of the tennis ball's weight W.

8 0
3 years ago
In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is appro
blondinia [14]
In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass. 
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were. 

Ok, Rameses:

Elementary charge . . . . .  1.6 x 10⁻¹⁹  coulomb
                                        negative on the electron
                                        plussitive on the proton

Electron rest-mass . . . . .  9.11 x 10⁻³¹  kg


a).  The force between two charges is

      F  =  (9 x 10⁹) Q₁ Q₂ / R²

          =  (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²

          =     ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²

          =          8.05 x 10⁻⁸  Newton .


b).  Centripetal acceleration  = 

                                               v² / r  .

                  A  =  (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)

                     =      7.7 x 10²²  m/s² .

That's an enormous acceleration ... about  7.85 x 10²¹  G's !
More than enough to cause the poor electron to lose its lunch.

It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use  "  F = MA "
either.

I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether  F = M A .

I'm going to leave that step for you.   Good luck !
4 0
3 years ago
A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately
Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car V_{c} = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

V_{f}  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

V_{f} =  V_{i} + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

d_{c} = V_{c} × t₁

we substitute

d_{c} = 22.352 × 7

d_{c}  = 156.46 m

now distance between polive car and speeding car

Δd =  d_{c} - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( V_{f} - V_{c} )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0
3 years ago
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