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3 years ago

The maximum gauge pressure in a hydraulic lift is 18.0 atm what is the largest size vehicle

1 answer:

3 0

There is a missing part in the question. Found the complete text on internet:
"What is the largest size vehicle (kg) it can lift if the diameter of the output line is 28.0 cm? "

Solution
The diameter of the piston is 28.0 cm, this means its radius is 14.0 cm (half the diameter), so the area of the piston is
 $A=\pi r^2 = \pi (0.14 m)^2 =6.15 \cdot 10^{-2} m^2$

The maximum pressure of the lift is
 $p=18.0 atm = 1.82 \cdot 10^6 Pa$

Therefore the maximum force the piston can lift is
 $F=pA=(1.82 \cdot 10^6 Pa)(6.15 \cdot 10^{-2} m^2)=1.12 \cdot 10^5 N$

And the size (the mass) of the vehicle is
 $m= \frac{F}{g}= \frac{1.12 \cdot 10^5 Pa}{9.81 m/s^2} =1.14 \cdot 10^4 kg$

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A 27.0 g marble sliding to the right at 56.8 cm/s overtakes and collides elastically with a 13.5 g marble moving in the same dir

Nataly [62]

Answer: $v_1=28.4 cm/s$

Explanation:

Given

mass of marble $m_1=27 gm$

velocity of marble $u_1=56.8 cm/s \approx 0.568 m/s$

mass of second marble $m_2=13.5 gm$

Velocity of second marble $u_2=14.2 cm/s \approx 0.142 m/s$

After collision 13.5 gm marble moves to the right at i.e. $v_2=71 cm/s$

Conserving momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$27\times 56.8+13.5\times 14.2=27\times v_1+13.5\times 71$

$1533.6+191.7=27\cdot v_1+958.5$

$27\cdot v_1=766.8$

$v_1=\frac{766.8}{27}=28.4 cm/s$

7 0

3 years ago

Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Dimas [21]

Answer:

Abby is standing (4.5^2 + 2.3^2)^1/2 from the far speaker

D2 = 5.05 m from the far speaker

The difference in distances from the speakers is

5.05 - 4.5 = .55 m (Let y be wavelength, lambda)

n y = 4.5

(n + 1) y = 5.05 for the speakers to be in phase at smallest wavelength

y = .55 m subtracting equations

f = v / y = 340 / .55 = 618 / sec should be the smallest frequency

8 0

3 years ago

A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o

Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m) ⇒ h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv² ⇒ d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0

2 years ago

Why the effect of gravitational force is more in liquid than in solid?

LekaFEV [45]

Gravitational force depends only on mass and distance, not on the state of matter.
The forces of attraction between molecules in matter are electromagnetic in nature, not gravitational.
These attractive forces are stronger in a solid than in a liquid than in a gas.
Gravitational forces between molecules is completely negligible compared to the em forces.

So, key answer is inter-molecular forces of solids is stronger than liquids.

4 0

3 years ago

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Rank the following in terms of increasing inertia:

Naddik [55]

C. A 1200kg car is going 15m/s

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3 years ago

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