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Kruka [31]
4 years ago
11

how do you solve this. Recklesss randy drives down the road at 48mi/h. Looking down to send a text to his gf he takes his eyes o

ff the road for 5.4 sec. what is his speed in ft per sec (1 mile=5280ft) and how far does the cr travel in the time it takes to send a text.
Physics
1 answer:
Nady [450]4 years ago
3 0

The whole problem here is convert units of [miles per hour]
into units of [feet per second].

(48 mile/hour) x (5,280 feet/mile) x (hour/3,600 sec) =  <em>70.4 ft/sec</em>

In 5.4 sec, the car travels  (5.4 x 70.4) =  <em>380.16 feet</em> .

That's how far back from the wreck they have to go
to find Randy's missing parts.


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riadik2000 [5.3K]

Answer:

They developed a test for poisons in the human tissues

Explanation:

Around the 19th century, Mathieu Orfila and Robert Christison developed important tests that identified poisons in human tissues.

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3 years ago
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If a 0.200-kilogram ball sits on a shelf 2.00 meters from the floor, how much mechanical energy (me) does it possess? More than
iogann1982 [59]

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Mechanical energy  = 3.92 J

exactly 3.92 j

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As we know that mechanical energy is sum of kinetic energy and potential energy of the system

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A centrifuge in a medical laboratory rotates at an angular speed of 3,650 rev/min. when switched off, it rotates through 48.0 re
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First of all we need to convert everything into SI units.

Let's start with the initial angular speed, \omega _i = 3650 rev/min. Keeping in mind that
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we have
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And we should also convert the angle covered by the centrifuge:
\theta = 48.0 rev= 48.0 rev \cdot  2 \pi  \frac{rad}{rev}=301.4 rad

This is the angle covered by the centrifuge before it stops, so its final angular speed is \omega_f =0.

To solve the problem we can use the equivalent of
2aS = v_f^2 -v_i^2
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2 \alpha \theta = \omega_f^2-\omega_i^2
And by substituting the numbers, we can find the value of \alpha, the angular acceleration:
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AP PHYSICS SYSTEM OF EQUATIONS
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Walking at a speed of 2.1 m/s, in the first 2 s John would have walked

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Take this point in time to be the starting point. Then John's distance from the starting line at time <em>t</em> after the first 2 s is

<em>J(t)</em> = 4.2 m + (2.1 m/s) <em>t</em>

while Ryan's position is

<em>R(t)</em> = 100 m - (1.8 m/s) <em>t</em>

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4.2 m + (2.1 m/s) <em>t</em> = 100 m - (1.8 m/s) <em>t</em>

(2.1 m/s) <em>t</em> + (1.8 m/s) <em>t</em> = 100 m - 4.2 m

(3.9 m/s) <em>t</em> = 95.8 m

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