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sergejj [24]
3 years ago
6

A hare and a tortoise compete in a race over a straight course 1.00 km long. The tortoise crawls at a speed of 0.200 m/s toward

the finish line. The hare runs at a speed of 8.00 m/s toward the finish line for 0.800 km and then stops to tease the slow-moving tortoise as the tortoise eventually passes by. The hare waits for a while after the tortoise passes and then runs toward the finish line again at 8.00 m/s. Both the hare and the tortoise cross the finish line at the exact same instant. Assume both animals, when moving, move steadily at their respective speeds. (a) How far is the tortoise from the finish line when the hare resumes the race? (b) For how long in time was the hare stationary?
Physics
1 answer:
vagabundo [1.1K]3 years ago
6 0

Answer:

a) 5 m from the finish line

b) 4875 s

Explanation:

distance = 1 km

tortoise speed = 0.2 m/s

speed of hare = 8 m/s

time take by hare t reach 0.8 km = \dfrac{800}{8} = 100 s

time taken by tortoise to travel 1 km = \dfrac{1000}{0.2} = 5000s

time taken by the hare to travel 0.2 km = \dfrac{200}{8} = 25 s

time taken by the tortoise to travel 0.2 km = \dfrac{200}{0.2} = 500s

distance traveled by tortoise in 25 second = 25 × 0.2 = 5 m

a) when tortoise is 5 m from the finish line then hare will resume the race.

b) the hare will be stationary for = 5000 - 125  = 4875 s

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The difference between the above velocities is that they exist in opposite direction of each other. or it can be said that they are negative vectors of each other.

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3 years ago
Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water
tangare [24]

Answer:

T_b=107.3784\ ^{\circ}C

Explanation:

Given:

  • thickness of the base of the kettle, dx=0.52\ cm=5.2\times 10^{-3}\ m
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  • temperature of the top surface of the kettle base, T_t=100^{\circ}C
  • rate of heat transfer through the kettle to boil water, \dot Q=0.409\ kg.min^{-1}
  • We have the latent heat vaporization of water, L=2260\times 10^3\ J.kg^{-1}
  • and thermal conductivity of aluminium, k=240\ W.m^{-1}.K^{-1}

<u>So, the heat rate:</u>

\dot Q=\frac{0.409\times 2260000}{60}

\dot Q=15405.67\ W

<u>From the Fourier's law of conduction we have:</u>

\dot Q=k.A.\frac{dT}{dx}

\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}

where:

A= area of the surface through which conduction occurs

T_b= temperature of the bottom surface

15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}

T_b=107.3784\ ^{\circ}C is the temperature of the bottom of the base surface of the kettle.

6 0
4 years ago
At a certain location, a gravitational force with a magnitude of 350 newtons acts on a 70.-kilogram astronaut. What is the magni
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3. 5.0N/kg

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4 years ago
A 6.0 m wire with a mass of 50 g, is under tension. A transverse wave, for which the frequency is 810 Hz, the wavelength is 0.40
MrRissso [65]

Answer:

a) t = 0.0185 s = 18.5 ms

b) T = 874.8 N

Explanation:

a)

First we find the seed of wave:

v = fλ

where,

v = speed of wave

f = frequency = 810 Hz

λ = wavelength = 0.4 m

Therefore,

v = (810 Hz)(0.4 m)

v = 324 m/s

Now,

v = L/t

where,

L = length of wire = 6 m

t = time taken by wave to travel length of wire

Therefore,

324 m/s = 6 m/t

t = (6 m)/(324 m/s)

<u>t = 0.0185 s = 18.5 ms</u>

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b)

From the formula of fundamental frquency, we know that:

Fundamental Frequency = v/2L = (1/2L)(√T/μ)

v = √(T/μ)

where,

T = tension in string

μ = linear mass density of wire = m/L = 0.05 kg/6 m = 8.33 x 10⁻³ k gm⁻¹

Therefore,

324 m/s = √(T/8.33 x 10⁻³ k gm⁻¹)

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<u>T = 874.8 N</u>

8 0
3 years ago
022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati
svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

v_f = 24.5 m/s

Explanation:

Part 1)

initial speed of the ball upwards

v_i = 24.5 m/s

so maximum height of the ball is given by

H = \frac{v_i^2}{2g}

H = \frac{24.5^2}{2(9.80)}

H = 30.6 m

Part 2)

As we know that final speed will be zero at maximum height

so we will have

v_f - v_i = at

0 - 24.5 = (-9.8)t

t = 2.5 s

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

v_f = 24.5 m/s

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

y = \frac{1}{2}at^2

4.88 = \frac{1}{2}a(1^2)

a = 9.76 m/s^2

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

a = 90 m/s^2

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

v_f = v_i + at

v_f = 0 + (90)(0.33)

v_f = 29.7 m/s

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

y = \frac{1}{2}gt^2

y = \frac{1}{2}(9.81)(2.3^2)

y = 25.95 m

Part 2)

Distance traveled by it in horizontal direction is given as

d = v_x t

d = 2.92 \times 2.3

d = 6.72 m

6 0
3 years ago
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