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trapecia [35]
3 years ago
13

An athlete is running a 400m race around a 400m track. On the backstretch the athlete's velocity is 8m/s but he is running into

a 2m/s headwind. How large is the drag force that acts on him? Assume that the density of the air is 1.2kg/m^3, his cross sectional area is 0.5m^2, and the coefficient of drag is 1.1.
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

Answer:

33 N

Explanation:

v = Velocity of fluid = 8+2 = 10 m/s

\rho = Density of fluid = 1.2 kg/m³

C = Coefficient of drag = 1.1

A = Cross sectional area = 0.5 m²

Drag force is given by

F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 1.1\times 0.5\times (8+2)^2\\\Rightarrow F=33\ N

The drag force on the athlete is 33 N

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Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

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Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
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a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

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v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

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v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

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Goryan [66]

Answer:

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Speed of the wave is given as

v = \frac{d}{t}\\v = \frac{4.21}{8}\\v = 0.53 ms^{-1}

\lambda = wavelength of the harmonic wave

wavelength of the harmonic wave is given as

\lambda = \frac{v}{f} \\\lambda = \frac{0.53}{1.12} \\\lambda = 0.47 m

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