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skad [1K]
3 years ago
9

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

Physics
1 answer:
victus00 [196]3 years ago
3 0
There is a missing part in the question. Found the complete text on internet:
"<span>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
</span>A=\pi r^2 = \pi (0.14 m)^2 =6.15 \cdot 10^{-2} m^2
<span>
The maximum pressure of the lift is
</span>p=18.0 atm = 1.82 \cdot 10^6 Pa
<span>
Therefore the maximum force the piston can lift is
</span>F=pA=(1.82 \cdot 10^6 Pa)(6.15 \cdot 10^{-2} m^2)=1.12 \cdot 10^5 N
<span>
And the size (the mass) of the vehicle is
</span>m= \frac{F}{g}= \frac{1.12 \cdot 10^5 Pa}{9.81 m/s^2} =1.14 \cdot 10^4 kg<span>
</span>
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Explanation:

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7 0
3 years ago
The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How
Vlada [557]

Answer:

The value is  w =  7.54 \  m        

Explanation:

From the question we are told that

     The length of the crack is  a =  0.3 \  m

     The  frequency is  f =  30.0 \ kHz =  30 *10^{3} \  Hz

      The distance outside the cave that is being consider is  D =  100 \  m

      The speed of sound is v_s =  340 \  m/s

Generally the wavelength of the wave is mathematically represented as

        \lambda =  \frac{v}f}

=>     \lambda =  \frac{340 }{30*10^{3}}

=>     \lambda = 0.0113 \ m/s

Generally for a  single slit the path difference between the interference patterns of the sound wave and the center  is mathematically represented as  

          y =  \frac{ n *  \lambda * D}{a}

=>     y =  \frac{ 1  *  0.0113 * 100}{0.3}

=>     y = 3.77 \  m

Generally the width of the sound beam is mathematically represented as

         w =  2 *  y

=>      w =  2 *  3.77

=>      w =  7.54 \  m        

4 0
3 years ago
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ValentinkaMS [17]

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Explanation:

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Time for total flight,

t = \frac{2vsin \theta}{g}

t'=\frac{vsin \theta}{g} = \frac {20 m/s}{9.8 m/s^2} = 2.04 s

Thus, the football takes 2.04 s to rise to the highest point of its trajectory.

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