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larisa86 [58]
3 years ago
6

A biker goes up hill with a constant speed of 10km/h. Then he goes downhill with a constant speed of 50km/h. What's his average

speed
Physics
1 answer:
Sladkaya [172]3 years ago
4 0

This is not as simple as it looks.  

His average speed is NOT (10km/hr + 50km/hr)/2 = 30 km/hr.

You have to use the definition of speed:

Speed = (total distance covered) / (time to cover the distance).

Let's say the distance up (and down) the hill is 'd' .

Then the time it takes to go up the hill is (d/10) hours.

And the time it takes to come down the hill is (d/50) hours.

Total distance = 2d km

Total time = (d/10) + (d/50) = (5d/50) + (d/50) = 6d/50

Speed = distance/time = 2d/(6d/50) = 100d/6d

<em>Speed = </em>100/6 = <em>16-2/3 km/hr</em>

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I’d say at 1
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3 years ago
Hi, so i have to find T1, can some1 help?
iragen [17]

30.1 N

Explanation:

Given:

W_1 = 16\:\text{N}

W_2 = 8\:\text{N}

Let's write the components of the net forces at the intersections. Note that the system is equilibrium so all the net forces are zero.

<u>Forces</u><u> </u><u>involving</u><u> </u><u>W1</u><u>:</u>

x:\:\:\:-T_1 + T_3\cos \alpha = 0\:\: \\ \text{or}\:\:T_2 = T_3\cos \alpha\:\:\:\:\:(1)

y:\:\:\:T_3\sin \alpha - W_1 = 0\:\:\: \\ \text{or}\:\:\:T_3\sin \alpha = W_1\:\:\:\:\:\:(2)

<u>Forces</u><u> </u><u>involving</u><u> </u><u>W2</u><u>:</u>

x:\:\:\:T_1\sin 53 - T_3\sin \alpha = W_2\:\:\:\:\:\:\:(3)

y:\:\:\:T_4 - T_1\cos 53 - T_3\cos \alpha = 0\:\:\:\;(4)

Substitute (2) into (3) and we get

T_1\sin 53 - W_1 = W_2

Solving for T_1,

T_1 = \dfrac{W_1 + W_2}{\sin 53} = 30.1\:\text{N}

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3 years ago
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This measure of the total amount of matter an object contains is called mass. The unit used for mass is the kilogram or kg for short.

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