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Leona [35]
1 year ago
5

The small piston of a hydraulic press has an area of 3.2 in2. If the applied force is 2.0 lb, what must the area of the large pi

ston be to exert a pressing force of 795 lb?
Physics
1 answer:
BartSMP [9]1 year ago
7 0

Given data:

The area of the hydraulic press is,

A_1=3.2in^2

The applied force is,

F_1=2\text{ lb}

The pressing force on the large piston is

F_2=795\text{ lb}

Given hydraulic press works on Pascal's law principle.

From the Pascal's law, the equation can be given as,

\frac{F_1}{A_1}=\frac{F_2}{A_2}

Here,

A_2

is the area of the larger piston.

Substituting the values in the above equation, we get:

\begin{gathered} \frac{2\text{ lb}}{3.2in^2}=\frac{795\text{ lb}}{A_2} \\ A_2=1272in^2 \end{gathered}

Thus, the area of the large piston is

1272in^2

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The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

5 0
2 years ago
A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below.
Alika [10]

Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:

    B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

    C) The maximum force is: F = 24 N

    D) The time to stop the block is: t = 25 s

 

Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.

B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.

x-axis

      -10 + 12 sin 60 + Fₓ = 0

        Fₓ = 10- 12 sin 60 = -0.39 N

y-axis

       12 cos 60 - 6 + F_y = 0

        F_y = 6 - 12 cos 60 = 0 N

We can give the result of the force in two ways:

  • Form of coordinates F = -0.39 i ^ N
  • Form of module and angle.

Let's use Pythagoras' theorem to find the modulus.

       F = \sqrt{F_x^2 + F_y^2 } \\F = \sqrt{0.39^2 +0^2}  

       F = 0.39N

We use trigonometry for the angle.

       tan \theta = \frac{F_y}{F_x}

       tan θ=  0º

The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.

        θ = 180 + θ'

        θ = 180 + 0

        θ = 180º

C) if the three forces can be moved and the maximum force occurs when they are all linear.

          10+ 6 + 6 + F = 0

          F = -24 N

D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.

          a = \frac{F}{m}  

          a = \frac{24}{6}  

         a =  4 m / s²

The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s

If we use kinematic relations.

        v = v₀ - a t

Final velocity when stopped is zero

         t = \frac{0-v_o}{a}

         t = 100/4

         t = 25 s

In conclusion using Newton's second law and the kinematics relations we can find the results for the questions about the forces and the motion of the block are:

    B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

    C) The maximum force is: F = 24 N

    D) The time to stop the block is: t = 25 s

Learn more about Newton's second law here: brainly.com/question/25545050

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The threatened or actual use of physical force to get someone to have sex without consent is called
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Wouldn't it be r a p e?
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Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from a
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Answer:

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