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3 years ago

Kiley went 5.7 km/h north and then went 5.8 km/h west. From start to finish, she went 8.1 km/h northwest.

1 answer:

6 0

Yes, that's a reasonable rounded value for the solution.

   (5.7)² + (5.8)² = (32.49) + (33.64) = 66.13

   √66.13 = 8.132...

   = 8.1 when rounded to the nearest tenth.

The solution ' 8.1 ' is a reasonable rounded value, but only
if the question is changed to say 'km' at every place where
it now says 'km/hr'.

If 'km/hr' is correct, then there's no way to calculate Kiley's
effective northwesterly speed, using only the given information.

We don't know how long she traveled north at 5.7 km/hr,
and we don't know how long she traveled west at 5.8 km/hr.
So we don't know the distance between her start and end
points, and we don't know how long she traveled altogether ...
exactly the two numbers we need in order to calculate her
average speed. Or even, for that matter, the average direction
of her trip from start to finish.

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2 years ago

Question 14 (1 point)

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3 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

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1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

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3 years ago

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Answer:

c. A number and a unit

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3 years ago

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Answer:

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