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liberstina [14]
3 years ago
10

A tortoise and a hare engage in a race. A tortoise can run with a speed of 0.15 m/s. A hare can run 25 times as fast as the tort

oise. In the race, they start at the same time; however, the hare being very proud of his much higher speed stops for a nap for 5.0 minutes. The tortoise wins the race by 35 cm. (a) What is the length of the race? (b) How long does the race take?
Physics
2 answers:
Ronch [10]3 years ago
8 0

Answer:

the answer is 30 minutes

Explanation: its on edge answers

horrorfan [7]3 years ago
5 0

Answer:

Explanation:

The speed of hare = .15 x 25 = 3.75 m /s . Let tortoise took t second to complete the race .

Distance traveled by it = .15 t

Distance traveled by hare = .15 t - .35 m

Time taken by hare to complete this distance

=  t - 5 x 60 s

Speed of hare

= Distance / time

( .15t-.35 ) / t - 300 , so

\frac{.15t-.35}{t-300} = 3.75

t = 312.40

= 5 minutes 12.4 seconds

Distance of race

312.4 x speed of tortoise

= 312.4 x .15

= 46.85 m

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Alexxandr [17]

Answer:

<em>vf=79.38 m/s</em>

<em>y= 321.5 m</em>

Explanation:

<u>Free Fall Motion</u>

A free-falling object falls under the exclusive influence of gravity. Free-falling objects do not encounter air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is g = 9.8 m/s^2.

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The distance traveled by a dropped object is:

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

1.

The penny will fall for t=8.1 s before hitting the ground, thus the height from which it was dropped is:

\displaystyle y=\frac{9.8\cdot 8.1^2}{2}

y= 321.5 m

2.

The final velocity is:

vf=9.8\cdot 8.1

vf=79.38 m/s

8 0
3 years ago
The early experiments with static charges were done by:
Naddika [18.5K]

Answer:

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Explanation:

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3 years ago
Your car's 30.0 W headlight and 2.50 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would
Tatiana [17]

Answer:

<h3>The power of headlight in series connection is 29.64 W</h3>

Explanation:

Given :

Power of headlight P_{1} = 30 W

Power of starter P_{2} = 2500 W

Voltage of headlight and starter V = 12 V

From equation of power,

 P = \frac{V^{2} }{R}

 R = \frac{V^{2} }{P}

For finding the resistance of headlight and starter,

⇒ For headlight,

 R_{1}  = \frac{144}{30} = 4.8 Ω

⇒ For starter,

R_{2} = \frac{144}{2500} = 0.057 Ω

Since equivalent resistance,

R_{eq} = R_{1} + R_{2} + ........

R_{eq} = 4.8 +0.057 = 4.857 Ω

So power in series is given by,

 P_{s } = \frac{V^{2} }{R_{eq} }  = \frac{144}{4.857}

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8 0
4 years ago
In root pass welding, that electrode we are going to use?​
postnew [5]

Answer:

The root pass is made with a 5/32” (4.0mm) diameter electrode. A cellulosic electrode (E-XX10) is being used. The root pass is welded with reverse (DC+) polarity.

Explanation:

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3 years ago
Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t
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Answer:

\frac{D_{jet}}{D_{prop}}=2.865

Explanation:

Given data

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Speed of prop driven Vprop=595 km/h

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Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

D=\frac{1}{2}CpAv^2\\

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

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4 0
3 years ago
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