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Leona [35]
2 years ago
8

in a hydraulic press the small cylinder has a diameter 10.0cm while the large has 25cm if the force of 600N is applied to the sm

all cylinder. find the force exacted on the large cylinder
Physics
2 answers:
Butoxors [25]2 years ago
8 0

Answer:

3750 N

Explanation:

Pressure on the small cylinder = pressure on the large cylinder

P₁ = P₂

F₁ / A₁ = F₂ / A₂

F₁ / (π d₁² / 4) = F₂ / (π d₂² / 4)

F₁ / d₁² = F₂ / d₂²

600 N / (10.0 cm)² = F / (25.0 cm)²

F = 3750 N

oee [108]2 years ago
8 0

Answer:

3751.34N

Explanation:

Pressure in large piston = pressure in smaller piston

P2 = P1

Pressure = Force / Area

Area = pi * r²

r1= d1/2 = 10/ 2 = 5cm = 0.05m

Area(A1) = 22/7 * (0.05)² = 0.00785m²

r2 = d2 / 2 = 25/2 = 12.5cm = 0.125m

Area(A2) = 22/7 * (0.125)² = 0.04908m²

Pressure = Force / Area

F1/A1 = F2/A2

600 / 0.00785 = F2 / 0.04908

F2 = (600 * 0.04908) / 0.00785

F2 = 3751.34N

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5 0
1 year ago
A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radiu
Pepsi [2]

Answer:

v = 10.89\ m/s

Explanation:

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radius of loop = 12.1 m                              

to find the minimum speed transverse by the rider to not to fall out upside down                                                                

centripetal force = \dfrac{mv^2}{r}

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5 0
2 years ago
one horsepower is a unit of power equal to 746w. how much energy can a 150-horsepower engine transform in 10.0s?
Dafna1 [17]

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

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=  (150 x 746 x 1 x 10)  joule  =  1,119,000 joules .   
if correct plz mark brainly
8 0
1 year ago
A cylinder of radius R and height H is floating upright in
emmainna [20.7K]

Answer:

Pressure difference between Top and Bottom of the cylinder is given as

\Delta P = \frac{gH}{2}(\rho_A + \rho_B)

Explanation:

As we know that the force due to pressure is balanced by the weight of the cylinder

So we will have

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so we have

\Delta P \pi R^2 = mg

so we have

\Delta P \pi R^2 = \pi R^2(\rho_A(\frac{H}{2}) + \rho_B(\frac{H}{2}))g

so we have

\Delta P = \frac{gH}{2}(\rho_A + \rho_B)

6 0
2 years ago
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