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My name is Ann [436]
3 years ago
14

The cheetah is one of the fastest accelerating animals, for it can go from rest to 28.0 m/s in 5.20 s. If its mass is 100 kg, de

termine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in (a) watts and (b) horsepower.
Physics
1 answer:
Setler79 [48]3 years ago
6 0

Answer:

a)15077 W

b)20.2185 horse power

Explanation:

P=F*V

F=ma

a=Vf-VS/t

Vf=28m/s

t=5.2

a=28/5.2

a=5.384 m/s²

F=100kg*5.384m/s²

F=538.4 N

P=F*V

P=538.4N*28m/s

P=15077 W=20.2185 horse power

1W=0.00134 Horse power

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Explanation:

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The anemometer is spinning and the pressure us high
skelet666 [1.2K]

Answer:

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2 years ago
A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit
Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c =  3 * 10⁸ Hz

Wavelength of the incident wave in air:

\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3

\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

n_1 = \sqrt{\frac{4\pi * 10^{-7}  }{8.84 * 10^{-12}  } }

n_1 = 377 \Omega

The intrinsic impedance of media 2 is given as:

n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

ϵr = 9

n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }

n_2 = 125.68 \Omega

c) The reflection coefficient,r  and the transmission coefficient,t at the boundary.

Reflection coefficient, r = \frac{n - n_{0} }{n + n_{0} }

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

r = \frac{3 - n_{0} }{3 + n_{0} }

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is E_{0} = 10 V/m

Maximum amplitudes in the total field is given by:

E = tE_{0} and E = r E_{0}

E = 10r, E = 10t

3 0
3 years ago
How much tension must a rope withstand if used to accelerate a 960-kg car from rest horizontally along a frictionless surface to
Svetach [21]
Use a=(dv/dt)         (change in velocity/ change in time)=acceleration
(1.2/5)=acceleration

F=ma (Newton's second law, Force= Mass x Acceleration

=960 x 0.24     F=230.4N If  T<230.4N then the tow rope will hold

7 0
3 years ago
When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play
saw5 [17]

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

6 0
3 years ago
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