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How much power is required to carry a 35 N package a vertical distance of 18 m if the work on the package is accomplished in 30
1 answer:
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Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
= rω²
ω² = / r
ω = √( / r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a = g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849
The power require to keep the car traveling is 6,666 W.
The power of the engine at the given efficiency is 3,999.6 W.
<h3>What is Instantaneous power?</h3>This the product of force and velocity of the given object.
The power require to keep the car traveling is calculated as follows;
P = Fv
The power of the engine at the given efficiency is calculated as follows;
Learn more about efficiency here: brainly.com/question/15418098