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1 year ago

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Sandy is on a road trip. She leaves at 8:00 AM. It takes her 2 hours to drive 200 kilometers. She stops at a rest stop for half

an hour. Then, she drives for 100 more kilometers, which takes her an hour and a half. What is Sandy's average velocity?

1 answer:

lutik1710 [3]1 year ago

0 0

The average velocity of Sandy is given by the total distance covered S divided by the total time taken t:
$v= \frac{S}{t}$

The total distance covered is
$S=200 km+0+100 km=300 km$
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
$t=2h+0.5h+1.5 h=4 h$
Therefore, the average velocity is
$v= \frac{S}{t}= \frac{300 km}{4 h}=75 km/h$

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<span> (a) does an electric field exert a force on a stationary charged object?
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As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

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Yes, The intensity of the electric force is still
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<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
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And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
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1 year ago

An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo

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Answer:

24.71 mm

Explanation:

Distance is proportional to focal length, so

d∝f

which means

$\frac{d'_1}{d'_2}=\frac{f_1}{f_2}$

Magnification of first lens

$M_2=-\frac{d'_1}{d_1}$

and

$M_2=\frac{h'_1}{h_1}$

Similarly, magnification of second lens

$M_2=-\frac{d'_2}{d_1}$

and

$M_2=\frac{h'_2}{h_1}$

From the above equations we get

$\frac{M_1}{M_2}=\frac{d'_1}{d_2'}$

and

$\frac{M_1}{M_2}=\frac{h'_1}{h_2'}$

which means,

$\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}$

and

$\frac{d'_1}{d_2'}=\frac{f_1}{f_2}$

So, we get

$\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm$

∴ Focal length should this camera's lens is 24.71 mm

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1 year ago

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Answer:

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Explanation:

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By conservation of energy:

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2 years ago

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