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3 years ago

11

A hydraulic lift raises a 2 000-kg automobile when a 500-N force is applied to the smaller piston. If the smaller piston has an

area of 10 cm2, what is the cross-sectional area of the larger piston?

1 answer:

Marina CMI [18]3 years ago

6 0

Answer:

Explanation:

ratio of the forces is equal to the ratio of the surface areas

which is another way of saying the pressure is constant in the system

2000(9.8) / 500 = A / 10

A = 2000(9.8)(10) / 500 = 392 cm²

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What would the final volume of 40 L of gas at 80 pascals be if the pressure increases to 130 pascals?

melomori [17]

Answer:

Final volume, V2 = 24.62 L

Explanation:

Given the following data;

Initial volume = 40 L

Initial pressure = 80 Pa

Final pressure = 130 Pa

To find the final volume V2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

$PV = K$

$P_{1}V_{1} = P_{2}V_{2}$

Substituting into the equation, we have;

$80 * 40 = 130V_{2}$

$3200 = 130V_{2}$

$V_{2} = \frac {3200}{130}$

$V_{2} = 24.62$

Final volume, V2 = 24.62 Liters

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3 years ago

Fossil A is found in a rock layer above a layer containing Fossil B. Which fossil is probably older? A) Fossil A B) Fossil B C)

luda_lava [24]

Fossil B is older the lower they found the fossil the older it is

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3 years ago

Read 2 more answers

A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

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3 years ago

A car is traveling on the highway at 28 m/s The driver sees a tree in the middle of the road and slams on the brakes. The car co

wel

Answer:what are the answer options?

Explanation

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3 years ago

You need to repair a gate on the farm. The gate weighs 100 kg and pivots as indicated. A small diagonal bar supports the gate an

tekilochka [14]

Answer:

The force is $F = 3920 \ N$

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The weight of the gate is $G = 100\ kg$

The vertical component of F is $F_y = F\ sin \theta$

From the diagram , taking moment about the pivot we have

$W_g * 2 - F_y * 1 = 0$

Where $W_g$ is the weight of the gate evaluated as

$W_g = m_g * g = 100 * 9.8 = 980 \ N$

=> $980 * 2 - Fsin(30) * 1 = 0$

=> $F = \frac{1960}{sin(30)}$

=> $F = 3920 \ N$

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3 years ago