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2 years ago

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A car travels from Bangalore to Mumbai and returns back on 20 hrs. What is the average velocity and average speed of the car if

odometer reading of the car, when it started was 2200 km and when finished was 3800 km?

2 answers:

irinina [24]2 years ago

7 0

Answer:

Explanation:

Average speed= Total distance/ Total time

Initial reading= 2200

final reading= 3800

total distance= Final reading -initial reading = 3800-2200 = 1600 km

total time= 20 hrs

Average speed= 1600/20

= 80 km/hr

CONVERTING km/hr TO m/s

To convert km/hr to m/s just multiply

Recall that,

1000m = 1 km

60 secs = 1 min

60 mins = 1 hr

( 80 × 1000 ) meters / ( 60×60 ) secs

80000m / 3600s

= 22.22m/s

Shtirlitz [24]2 years ago

4 0

Answer:

Average speed = 80 km/hr

Average velocity = 0 km/hr

Explanation:

Average velocity is given by

Average velocity = Displacement / Time taken

And Average speed is given by

Average speed = Total distance / Time taken

To calculate the average speed, we will first find the total distance,

Total distance = Final reading of the odometer - Initial reading of the odometer

Hence,

Total distance = 3800km - 2200km

Total distance = 1600km

From the question, the time taken = 20hrs

Therefore,

Average speed = 1600km ÷ 20hrs

Average speed = 80 km/hr

To calculate the average velocity, we will find the displacement.

Displacement is the distance traveled in a straight line OR the shortest distance from the initial to the final position.

Since the car returned back to its initial position (Bangalore), the displacement is zero.

From,

Average velocity = Displacement / Time

Hence,

Average velocity = 0 ÷ 20

Average velocity = 0 km/hr

The average velocity is 0 km/hr

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A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w

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Answer:

The dog catches up with the man 6.1714m later.

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Now, the distance equation for the man would be:

$d_{man}=v_{man}\cdot t=1.6\cdot t$

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

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$d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0$

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

$d_{man}=d_{dog}$

$1.6\cdot t=3\cdot (t-1.8)$

$1.6\cdot t=3\cdot t-5.4$

$1.4\cdot t=5.4$

$t=\frac{5.4}{1.4}$

$t=3.8571s$

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

$d_{man}=1.6\cdot t=1.6\cdot (3.8571)$

$d_{man}=6.1714m$

Finally, the dog catches up with the man 6.1714m later.

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To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as $V_1$ , the relation with the final volume as

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Let T be the temperature after expanding by the formula of volume expansion

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Where $\gamma$ is the volume coefficient of copper $5.1*10^{-5}/C$

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$1.00163 = 1+5.1*10^{-5}(T-21.1\°)$

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