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Marysya12 [62]
3 years ago
11

What is the correct forward position for making a right turn from a stopped position?

Physics
1 answer:
zysi [14]3 years ago
5 0
The correct forward position for making a  right turn from a stopped position is :
LP5
People should also do this when approaching an intersection where the lane opens up

hope this helps
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Speed1=20 m/s. time1=5sec
lina2011 [118]

Answer:

<h2><em><u>2</u></em><em><u>1</u></em><em><u>.</u></em><em><u>2</u></em><em><u>5</u></em><em><u> </u></em><em><u>m</u></em><em><u>/</u></em><em><u>s</u></em></h2>

Explanation:

<h2><em>AVERAGE</em><em> </em><em>S</em><em>P</em><em>E</em><em>E</em><em>D</em><em>=</em></h2>

<em>=  \frac{s1 t1+ s2t2 + s3t3}{t1 + t2 + t3}</em>

<em>=  \frac{20 \times 10 + 30 \times 5 + 15 \times 5}{10 + 5 + 5}</em>

<em>=  \frac{200 + 150 + 75}{20}</em>

<em>=  \frac{425}{20}</em>

<em>= 21.25 ms</em>

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2 years ago
Which of these instruments can be used to peedict a thunderstorm?
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<span> Doppler Radar is the instrument that is used to predict a thunderstorm. </span>
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3 years ago
) The potential difference between the plates of a capacitor is 400 V. (a) If the spacing between the plates is doubled without
kramer

(a) 800 V

The capacitance of a parallel-plate capacitor is given by

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the separation between the plates

We see that the capacitance is inversely proportional to the separation, d: in this problem, the separation between the plates is doubled (2d), so the capacitance will become half of its original value:

C'=\frac{C}{2}

The potential difference between the plates is related to the capacitance by

V=\frac{Q}{C} (1)

where Q is the charge stored on the plate. In this problem, the charge is not changed: therefore, the new potential difference is

V'=\frac{Q}{C/2}=2\frac{Q}{C}=2V

So, the potential difference has doubled, and since the initial  value was

V = 400 V

The new value is

V' = 2(400) = 800 V

(b) The charge will decrease by a factor 2

As before, the plate spacing is doubled, so according to the equation

C=\frac{\epsilon_0 A}{d}

Then the capacitance is halved again:

C'=\frac{C}{2}

This time, however, the voltage is held constant. We can rewrite the eq.(1) as

Q = CV

And since V has not changed, we can find what is the new charge stored in the capacitor:

Q' = C'V=\frac{C}{2}V=\frac{1}{2}(CV) = \frac{1}{2}Q

So, the charge will be halved compared to the original charge.

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3 years ago
phosphorus-32 undergoes beta decay. which particle correctly completes the equation to show that the numbers of nucleons on the
Alex17521 [72]

Answer:A

Explanation:

A P E X

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3 years ago
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When putting the ball on the tee you want half of the golf ball to be above the club be below the club be in front of the club b
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5)-5 that would be 100% correct
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