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Usimov [2.4K]
3 years ago
9

To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance

. Suppose a cyclist is traveling at 15 km/h on level ground. Assume he is using 480 W of metabolic power.
a. Estimate the amount of power he uses for forward motion.
b. How much force must he exert to overcome the force of air resistance?
Physics
1 answer:
mart [117]3 years ago
3 0

Answer:

a. 120 W

b. 28.8 N

Explanation:

To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance. Suppose a cyclist is traveling at 15 km/h on level ground. Assume he is using 480 W of metabolic power.

a. Estimate the amount of power he uses for forward motion.

b. How much force must he exert to overcome the force of air resistance?

(a) He is 25% efficient, therefore the cyclist will be expending 25% of his power to drive the bicycle forward

Power = efficiency X metabolic power

= 0.25 X 480

= 120 W

(b)

power if force times the velocity

P = Fv

convert  15 km/h to m/s

v = 15 kmph = 4.166 m/s

F = P/v

= 120/4.166

= 28.8 N

definition of terms

power is the rate at which work is done

force is that which changes a body's state of rest or uniform motion in a straight line

velocity is the change in displacement per unit time.

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Answer:

Frequency

Explanation:

The property of waves that remains unchanged as it crosses the boundary of one medium to another is the frequency of the wave.

As a wave moves from one boundary to another, the wavelength and the speed of the wave changes.

The speed of the wave is product of wavelength and frequency. Also, the wavelength of the wave is function of the distance between successive crests or troughs of a wave.

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A sprinf has a potential energy of 84.08 J and a constant of 342.25 N/m. How far it been stretched? Use potential energy elastic
valina [46]

The spring has been stretched 0.701 m

Explanation:

The elastic potential energy of a spring is the potential energy stored in the spring due to its compression/stretching. It is calculated as

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For the spring in this problem, we have:

E = 84.08 J (potential energy)

k = 342.25 N/m (spring constant)

Therefore, its elongation is:

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A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of
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The largest mass of cargo the balloon can lift is 791.06 kg

First, we need to calculate the mass of helium.

Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.

The volume of the balloon also equals the volume of helium present.

Now, the mass of helium m = density of helium, ρ × volume of helium, V

m = ρV

Since ρ = 0.179 kg/m³

m = ρV

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m = 0.179 kg/m³ × 4π(7.15 m)³/3

m = 0.179 kg/m³ × 4π(365.525875 m³)/3

m = 0.179 kg/m³ × 1462.1035π m³/3

m = 261.7165265π/3 kg

m = 822.207/3 kg

m = 274.07 kg

Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.

The weight of this mass W = Mg where g = acceleration due to gravity.

The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.

W' = m'g

Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'

We know that the volume of air displaced, V' = volume of balloon, V

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Since the density of air, ρ' = 1.29 kg/m³,

m' = ρ'V

m = 1.29 kg/m³ × 4π(7.15 m)³/3

m = 1.29 kg/m³ × 4π(365.525875 m³)/3

m = 1.29 kg/m³ × 1462.1035π m³/3

m = 1886.113515π/3 kg

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So, the net mass m" = W"/g = 791.06g/g = 791.06 kg

This net mass is the largest mass of cargo that the balloon can lift.

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marshall27 [118]

Answer:

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Unknown:

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