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Solnce55 [7]
3 years ago
7

Hydraulic systems utilize Pascal's principle by transmitting pressure from one cylinder (called the primary) to another (called

the secondary). Since the pressures will be equal, if the surface areas are different then the forces applied to the cylinders' pistons will be different. Suppose in a hydraulic lift, the piston of the primary cylinder has a 1.85-cm diameter and the piston of the secondary cylinder has a 24.5-cm diameter. What force, in newtons. must be exerted on the primary cylinder of this lift to support the weight of a 2200-kg car (a large car) resting on the secondary cylinder?
Physics
1 answer:
Paul [167]3 years ago
3 0

Answer:

F1= 122.93 N

Explanation:

Pascal´s Principle

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

P=F/A

P1=P2

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

Data

D1 =  1.85 cm : primary cylinder piston diameter

D2 = 24.5 cm : secondary cylinder piston diameter

m = 2200-kg  : car mass

Piston area calculation

A_{1} = \frac{\pi *(D_{1})^{2}  }{4}

A_{1} = \frac{\pi *(1.85)^{2}  }{4}

A1= 2.688 cm²

A_{2} = \frac{\pi *(D_{2})^{2}  }{4}

A_{2} = \frac{\pi *(24.5)^{2}  }{4}

A2 = 471.435 cm²

Calculating of the weight of the car (W)

W = m*g = 2200-kg * 9.8 m/s² = 21560 N

Calculation of the force in Newtons to be exerted on the primary cylinder piston

Data:

A1= 2.688cm²

A2= 471.435 cm²

F2 = W=  21560 N

We replace data in the formula (1)

\frac{F_{1} }{A_{1}} = \frac{F_{2} }{A_{2}}

F_{1} =\frac{F_{2}*A_{1}  }{A_{2} }

F_{1} =\frac{21560N*2.688 cm^{2} }{471.435 cm^{2} }

F1= 122.93 N

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alexgriva [62]

Answer:

202.95J

Explanation:

The formulae for the energy absorbed by the iron skillet is;

q=m*c*ΔФ   where q=energy absorbed, m is mass of iron skillet, c is specific heat capacity and ΔФ is change in temperature

Given that;

m= 41g

ΔФ=20°-9°=11°

c=0.45 J/g°C

q= 41×0.45×11

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4 0
3 years ago
If a star with an absolute magnitude of -5 has an apparent magnitude of +5 ,then its distance is
klio [65]
You asked a question.  I'm about to answer it. 
Sadly, I can almost guarantee that you won't understand the solution. 
This realization grieves me, but there is little I can do to change it. 
My explanation will be the best of which I'm capable.


Here are the Physics facts I'll use in the solution:

-- "Apparent magnitude" means how bright the star appears to us.

-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).

-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by  ⁵√100 .
That's about  2.512... .  

-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .

-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).

That's all the Physics.  The rest of the solution is just arithmetic.
____________________________________________________

-- The star in the question would appear M(-5) at a distance of
32.6 light years. 

-- It actually appears as a M(+5).  That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.

-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .

-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is

(32.6) · (100)^(1) light years

= (32.6) · (100) light years

=  approx.  3,260 light years .   (roughly 1,000 parsecs)


I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
6 0
3 years ago
A 645-turn coil with a 20.250 m ​2 ​​ area is spun in Earth’s 5.00×10 ​−5 ​​ T magnetic field, producing a 1.25-V maximum emf. A
Dmitriy789 [7]

Answer:

\omega = 1.914\ rad/s

Explanation:

Given,

Number of turns, N = 645 N

Area, A = 20.25 m²

Earth Magnetic field, B = 5 x 10⁻⁵ T

Maximum Emf = 1.25 V.

Angular velocity, ω = ?

Using Induced Emf formula

\varepsilon = NAB\omega

\omega= \dfrac{\varepsilon}{NAB}

\omega= \dfrac{1.25}{645\times 20.25\times 5\times 10^{-5}}

\omega = 1.914\ rad/s

Angular velocity of the coil = \omega = 1.914\ rad/s

5 0
3 years ago
An electron at Earth's surface experiences a gravitational force of magnitude F=(9.11×10−31 kg)⋅(9.8 m/s2). Part A How far away
katrin [286]

Answer:

r = 5,085 m

Explanation:

The force exerted by on the surface of the Earth on an electron is its weight

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          W = 8.9 10⁻³⁰ N

The electric force between an electron and a proton is given by Coulomb's Law

         Fe = k q₁ q₂ / r²

         Fe = - k q² / r²

They ask us that W = Fe

         W = k q² / r²

          r = √ k q² / W

Let's calculate

        r = √ 8.99 10⁹ (1.6 10⁻¹⁹)² /8.9 10⁻³⁰

        r = √ 25.86

        r = 5,085 m

Let's look for the relationship of this distance with the harmonic distance

        R / R_atomic = 5,085 / 10⁻¹⁰

        R / R_Atomic = 5 10¹⁰

We see that this distance is 10¹⁰ times the interatomic distance, so the gravitational attraction force is very small at atomic scale

8 0
3 years ago
A red ladybug appears in white light, in red light, and in blue light.
kirza4 [7]
<span>A red ladybug appears red in white light,red in red light, and black in blue light.</span>
3 0
4 years ago
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