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GalinKa [24]
2 years ago
6

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

a gazelle assumes an acceleration of 4.2m/s2
for 6.5 s, after which the gazelle continues continues at a steady speed.
a. What is the gazelle's top speed?
b. A human would win a very short race with a gazelle. The best time for a 30 m sprint for a human runner is 3.6 s. How much time would the gazelle take for a 30 m race?
c. A gazelle would win a longer race. The best time for a 200 m sprint for a human runner is 19.3s.
How much time would the gazelle take for a 200 m race?
Physics
1 answer:
Lyrx [107]2 years ago
4 0

Answer:

a) 27.3 m/s

b) 3.78 s

c) 10.576 s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

a)

v=u+at\\\Rightarrow v=0+4.2\times 6.5\\\Rightarrow v=27.3\ m/s

Top speed of the gazelle is 27.3 m/s

b)

s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0t+\frac{1}{2}\times 4.2\times t^2\\\Rightarrow t=\sqrt{\frac{30\times 2}{4.2}}\\\Rightarrow t=3.78\ s

The gazelle would take 3.78 seconds to win a 30 m race.

c)

s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 6.5+\frac{1}{2}\times 4.2\times 6.5^2\\\Rightarrow s=88.725\ m

The gazelle would go 88.725 m with the acceleration after which there will be no acceleration.

Time = Distance / Speed

\text{Time}=\frac{200-88.725}{27.3}=4.076\ s

Total time taken by the gazelle to cover 200 m would be 6.5+4.076 = 10.576 seconds

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Answer:

Explanation:

Given

Original Frequency f=723\ Hz

apparent Frequency f'=697\ Hz

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

f'=f\cdot \frac{v-v_o}{v+v_s}

where  

f'=apparent frequency  

v=velocity of sound in the given media

v_s=velocity of source

v_0=velocity of observer  

here v_0=0

697=723\cdot (\frac{343-0}{343+v_s})

v_s=(\frac{f}{f'}-1)v

v_s=(\frac{723}{697}-1)\cdot 343

v_s=12.79\approx 12.8\ m/s

i.e.fork acquired a velocity of 12.8 m/s

distance traveled by fork is given by

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v_s^2-0=2\times 9.8\times s

s=\frac{12.8^2}{2\times 9.8}

s=8.35\ m

                                       

5 0
2 years ago
After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r
mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

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2 years ago
Io, a satellite of Jupiter, is the most volcanically active moon or planet in the solar system. It has volcanoes that send plume
Mamont248 [21]

Answer:

1331.84 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement = 490 km

a = Acceleration

g = Acceleration due to gravity = 1.81 m/s² = a

From equation of linear motion

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -1.81\times 490000}\\\Rightarrow u=1331.84\ m/s

The speed of the material must be 1331.84 m/s in order to reach the height of 490 km

3 0
2 years ago
Read 2 more answers
A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s
sweet [91]

This question is in complete.The question is

A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.

Answer:

distance=0.124 m

Explanation:

mgh=mglSin\alpha =(1/2)Iw_{i}^{2}+(1/2)mv^{2}\\   v=wR\\Solve for L\\L=((1/2)(1/2)0.015^{2}*60^{2}+(1/2)(60*0.015^{2} ))/9.8Sin30\\   L=0.124m

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gladu [14]

Answer: 0.98m

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P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

8 0
2 years ago
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