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ziro4ka [17]
3 years ago
10

Consider a simple hydraulic system in an automotive shop. A worker can exert a force of 2000 N on a circular piston with an area

of 20 $cm^2$. Another piston is used to raise a car weighing 20000 N. What must the area of the second piston be in order to raise the car?
A. 20 cm^2B. 2000 cm^2C. 2 cm^2D. 200 cm^2
Physics
1 answer:
anastassius [24]3 years ago
8 0

Answer:

D) A₂ =  200 cm²

Explanation:

Pascal principle

When exerting a pressure on one of the vessels, this will be exerted with equal magnitude in all directions and in each part of the fluid, in this way the fluid will move towards the other container to achieve equilibrium again, that is, to equalize the pressures in both vessels.

The pressures must be the same in vessels 1 and 2, then:

P₁ = P₂ and therefore:

\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }   Equation (1)

Data

F₁ = 2000 N

A₁=  20 cm²

F₂= 20000 N

Calculating of the area of the second piston  (A₂)

We replace data in the equation (1)

\frac{ 2000  }{ 20 } = \frac{20000 }{A_{2} }  

100 = 20000/A₂

100 * A₂ = 20000

A₂ =  20000/100

A₂ =  200 cm²

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Electromagnetic wave allows us to control television with a remote control.

This electromagnetic wave is invisible.

Infrared is that electromagnetic wave that is used to control television using remote control.

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1 year ago
On a winter day with a temperature of -10°C, 500g of snow (water ice) is brought inside where the temperature is 18 °C. The snow
Serjik [45]

Answer:

Explanation:

Mass of ice m = 500g = .5 kg

Heat required to raise the temperature of ice by 10 degree

= mass of ice x specific heat of ice x change in temperature

= .5 x 2093 x 10 J

10465 J

Heat required to melt the ice

= mass of ice x latent heat

0.5 x 334 x 10³ J

167000 J

Heat required to raise its temperature to 18 degree

= mass x specific heat of water x rise in temperature

= .5 x 4182 x 18

=37638 J

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=10465 +167000+ 37638

=215103 J

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3 years ago
What two bodily functions are increased by a warm up
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6 0
3 years ago
How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium
Minchanka [31]

The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

<h3>Pressure and temperature at equilibrium </h3>

The relationship between pressure and temperature can be used to determine the height risen by the water.

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where;

  • V₁ = AL
  • V₂ = A(L - y)
  • P₁ = Pa
  • P₂ = Pa + ρgh
  • T₁ = 20⁰C = 293 K
  • T₂ = 10⁰ C = 283 k

\frac{PaAL}{T_1} = \frac{(P_a + \rho gh)A(L-y)}{T_2} \\\\\frac{PaL}{T_1} = \frac{(P_a + \rho gh)(L-y)}{T_2} \\\\L-y = \frac{PaLT_2}{T_1(P_a + \rho gh)} \\\\y = L (1 - \frac{PaT_2}{T_1(P_a + \rho gh)})\\\\y = 4.2(1 - \frac{101325 \times 283}{293(101325\  +\  1000 \times  9.8 \times  100)} )\\\\y = 3.8 \ m

Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

The complete question is below:

A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?

Learn more about thermal equilibrium here: brainly.com/question/9459470

#SPJ4

3 0
2 years ago
(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W
Kitty [74]

Answer:

The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

D\sin\theta=n\lambda

D=\dfrac{n\lambda}{\sin\theta}

Where, D = width of slit

\lambda = wavelength

Put the value into the formula

D=\dfrac{n\lambda}{\sin\theta}

Here, \sin\theta should be maximum.

So. maximum value of \sin\theta is 1

Put the value into the formula

D=\dfrac{1\times\lambda}{1}

D=\lambda

(b). If the minimum number  is 50

Then, the width is

D=\dfrac{50\times\lambda}{1}

D=50\lambda

(c). If the minimum number  is 1000

Then, the width is

D=\dfrac{1000\times\lambda}{1}

D=1000\lambda

Hence, The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

4 0
4 years ago
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