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nlexa [21]
3 years ago
14

a trampoline launches a 50kg person 2m into the air. if the springs push with 1960N of force, how much displacement was there in

the trampoline
Physics
2 answers:
Lina20 [59]3 years ago
6 0

Answer: 0.5 m

Explanation:

Given

Mass of the person is m=50\ kg

Trampoline launches the person into the air up to height of h=2\ m

Force experience by springs is F=1960\ N

Here, the work done on displacing the springs is equivalent to the Potential energy acquired by the person i.e.

\Rightarrow F\cdot x=mgh\quad [\text{x=displacement of the trampoline}]\\\\\text{Insert the values}\\\\\Rightarrow x=\dfrac{50\times 9.8\times 2}{1960}\\\\\Rightarrow x=\dfrac{980}{1960}\\\\\Rightarrow x=0.5\ m

DiKsa [7]3 years ago
3 0

Answer:

0.5

Explanation:

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Assume the population of a large city like New York City is about 4 × 10^6 people, with about 1.97 people per household. Approxi
Evgesh-ka [11]

So here are the given information:

4 x 10^6 people 

1.97 people = 1 household 

16.1 households = 1 piano 


1 piano = 1 tuning/year 

1 tuner = 3 tunings/working day 

<span>1 year = 250  working days </span>


Multiplying everything: 

4 x 10^6 people x (1 household/1.97 people) x (1 piano/16.1 households) = 126115.33 pianos in NYC 

126115.33 pianos x (1 (tuning/year)/1 piano) x (1 year/250 working days) x (1 working day/3 pianos/tuner) = 168.15 tuners 


<span>Therefore about 168 piano tuners in NYC.</span>

7 0
3 years ago
A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in
Gekata [30.6K]

A.

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

Substituting the values of the variables into the  equation, we have

Q = mcΔT

Q = ρVcΔT

Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C

Q = 482630400 J

So, the energy of the hot water is 482630400 J

B.

The elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

Substituting the values of the variables into the equation, we have

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Learn more about heat energy here:

brainly.com/question/11961649

5 0
2 years ago
The electric motor of a model train accelerates the train from rest to 0.540 m/s in 27.0 ms. The total mass of the train is 610
Gnesinka [82]

Answer:

The value  is  P  =3.294 \  W

Explanation:

From the question we are that

     The velocity  v  =  0.540  \  m/s

      The  time taken is  t =  27.0 ms  = 27.0 *10^{-3} \ s

     The  total mass of the train is  m  =  610 \ g  =  0.610 \ kg

Generally the average power delivered is mathematically represented as

     P  =\frac{KE }{t}    

     P  =\frac{  \frac{1}{2}  *  m  *  v^2 }{t}    

=>  P  =\frac{  \frac{1}{2}  *  0.610   *   0.540 ^2 }{ 27.0 *10^{-3}}    

=>  P  =3.294 \  W  

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3 years ago
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Some call it "air resistance", and others just call it "drag".

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