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salantis [7]
3 years ago
10

A. Group of cross country runners decided to go on an hour and a half run. During the first hour, they ran a total of 13 kilomet

ers. Then, they ran 5.0 kilometers during the next half an hour. What was the group's average speed for the entire run?
Physics
1 answer:
pshichka [43]3 years ago
3 0

Answer:

The average speed for the entire run is 12 km/h.

Explanation:      

The average speed is given by the following equation:

\overline{v} = \frac{d_{T}}{t_{T}}

Where:

d_{T}: is the total distance

t_{T}: is the total time

If during the first hour, they ran a total of 13 kilometers and then, they ran 5.0 kilometers during the next half an hour we have:

d_{T} = 13 km + 5 km = 18 km

t_{T} = 1 h + \frac{1}{2} h = 1.5 h

Hence, the average speed is:

\overline{v} = \frac{d_{T}}{t_{T}} = \frac{18 km}{1.5 h} = 12 km/h

Therefore, the average speed for the entire run is 12 km/h.

I hope it helps you!                                                                                      

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A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m
ArbitrLikvidat [17]

Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

\sum Fx = ma_x  \ and \ \sum Fy = ma_y

The magnitude of the net force which is also known as the resultant will be expressed as R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

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3 years ago
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34kurt
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A 1060-kg car moving west at 16 m/s collides with and locks onto a 1830-kg stationary car. determine the velocity of the cars ju
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therefore; 
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