The force that pulls the wagon is the horizontal component of the actual force, which is 290*cos(32) = 246N
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
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
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The maximum pressure of the lift is
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
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Therefore the maximum force the piston can lift is
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
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And the size (the mass) of the vehicle is
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
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Answer:
<h2>Magnetic field required for the given induced EMF is 1.41 T</h2>
Explanation:
Potential difference across the blood vessel is given as

here we know that the speed is given as



now we have


Now volume flow rate of the blood is given as


from above equation we have

Now we have


True because we have limited amount if it
Answer:
Yes.
Explanation:
Reactors use uranium for nuclear fuel. The uranium is processed into small ceramic pellets and stacked together onto sealed metal tubes called fuel rods. The heat created by fission turns the water into steam.