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Georgia [21]
3 years ago
11

A tortoise can run with a speed of 0.14 m/s, and a hare can run 20 times as fast. In a race, they both start at the same time, b

ut the hare stops to rest for 1.0 minutes. The tortoise wins by a shell (30 cm). (a) how long does the race take? S (b) what is the length of the race?
Physics
1 answer:
Reptile [31]3 years ago
4 0

speed of tortoise is given as v1 = 0.14 m/s

speed of hare is given as v2 = 20*0.14 = 2.8 m/s

now let say the total length of the path is "d"

so the total time taken by the tortoise to cover this

t = \frac{d}{0.14}

now given that hare took rest for 1 min

so total time of run for hare is (t - 60)s

so the distance that hare covered is given by

d - 0.30 = 2.8 * (t - 60)

now by above two equations

d - 0.30 = 2.8 * \frac{d}{0.14} - 168

168 - 0.30 = (20 - 1)d

d = 8.82 m

and the time t is given by

t = \frac{8.82}{0.14}

t = 63 s

so part a)

t = 63 s

part b)

d = 8.82 m

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Answer:

70.15 Joule

Explanation:

mass of man, m = 70 kg

intial length, l = 11 m

extension, Δl = 1.5 m

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A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f
Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

T=\dfrac{2\pi}{\omega}

{\omega}=\dfrac{2\pi }{0.2}\ rad/s

{\omega}=31.41\ rad/s

We know that

{\omega}^2=m\ K

K=Spring constant

K=\dfrac{\omega^2}{m}

K=\dfrac{31.41^2}{0.12}\ N/m

K=8221.56 N/m

The maximum force F

F= K A

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F=698.83 N

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3 years ago
Rama's weight is 40kg. She is carrying a load of 20 kg up to a height of 20 m . What work does she do?​
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Answer:

\huge\star{\underline{\mathtt{\blue{Answer}}}}\huge\star...

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.

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3 years ago
A jogger travels a route that has two parts. The first is a displacement of 3 km due south, and the second involves a displaceme
spayn [35]

Answer:

a) 2.41 km

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Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

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The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

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dy = \sqrt{d^2 - dx^2}

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The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

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