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Igoryamba
2 years ago
13

Driving Zone 7 corresponds with

Physics
1 answer:
stepan [7]2 years ago
3 0
Your driving zone refers to the areas of space around your car, it refers to all the area around your car as far as your eyes can see. 
Each car has seven zones numbered from 1 to 7. Driving zone 7 corresponds with THE SPACE YOUR VEHICLE IS OCCUPYING. The other zones are as follows:
zone 1 = area directly infront of your car
zone 2 = your left lane
zone 3 = your right lane
zone 4 = left rear of your car
zone 5 = right rear of your car 
zone 6 = area directly behind your car.
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A bowling ball encounters a 0.760 m vertical rise on the way back to the ball rack. Ignore frictional losses and assume the mass
Gre4nikov [31]

To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.

Let's start considering that the angular velocity is given by,

\omega = \frac{v}{R}

Where,

V = linear speed

R = the radius

In the case of the initial kinetic energy:

KE_i=\frac{1}{2} mv^2 + \frac{1}{2}I \omega^2

Where I is the moment of inertia previously defined.

KE_i = \frac{1}{2}(m)3.5^2 + \frac{1}{2}* (\frac{2}{5} m R^2) (\frac{3.5}{R})^2

In the case of the final kinetic energy, we have to,

KE_f= mgh+ \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2

KE_f = m * 9.81 * 0.76 + \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2

For conservation of Energy we have, that

KE_f = KE_i, then (canceling the mass and the radius)

\frac{1}{2} 3.5^2 + \frac{1}{2}(\frac{2}{5})(3.5)^2= 9.81 * 0.76 + \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2

8.575= 7.4556+ \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2

1.1194= \frac{1}{2}( v^2 + (\frac{2}{5}) (v)^2)

2.2388= (\frac{7}{5}) (v)^2

v=1.26m/s

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Answer:

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<em><u>STRATOSPHERE</u></em><em><u> </u></em><em>constant</em><em> </em><em>temperature</em><em> </em>

<em>MESOSPHERE</em><em> </em><em>layer</em><em> </em><em>with</em><em> </em><em>the</em><em> </em><em>strongest</em><em> </em><em>wind</em><em> </em><em>and</em><em> </em><em>below</em><em> </em><em>the</em><em> </em><em>themosphere</em>

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AURORKA [14]
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Convert 5 min to seconds and divide 2150J by that number.
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damaskus [11]
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Where mg represents the weight reading without external acceleration and ma as the weight reading due to the external acceleration.
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Please help with this! The selected answer is the incorrect one, I'm making corrections.
elixir [45]

Answer:

The answer is C-Spin a magnet around the wire on the piece of iron. Hope that helps :)

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