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2 years ago

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A bicyclist of mass 68 kg rides in a circle at a speed of 3.9 m/s. If the radius of the circle is 6.5 m, what is the centripetal

force on the bicyclist?A.41 N B.159 N C.179 D.92 N

1 answer:

ASHA 777 [7]2 years ago

3 0

Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m

Formula:
$F_{centripetal\:force} = \frac{m*s^2}{R}$

Solving:
$F_{centripetal\:force} = \frac{m*s^2}{R}$
$F_{centripetal\:force} = \frac{68*3.9^2}{6.5}$
$F_{centripetal\:force} = \frac{68*15.21}{6.5}$
$F_{centripetal\:force} = \frac{1034.28}{6.5}$
$\boxed{\boxed{F_{centripetal\:force} = 159.12\:N}}$
Answer:
<span>B.159 N</span>

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Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

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