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Monica [59]
4 years ago
9

Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 46.0 mph . Beth leaves Los A

ngeles at 9:00 a.m. and drives a steady 55.0 mph .How long does the first to arrive have to wait for the second?
Physics
1 answer:
elena-14-01-66 [18.8K]4 years ago
4 0

Answer:

25 min, 48 sec

Explanation:

Alan:

t = d/v

  = 400 mi / 46 mph = 8.70 hr

So it took Alan 8.70 hrs; from 8 am, that's 4:42 pm that he arrives.

t = d/v

  = 400 mi / 55 mph = 7.27 hr

So it took Beth 7.27 hrs; from 9 am, that's 4:16:12 that she arrives.

4:42 - 4:16:12 = 25.8 minutes = 25 minutes, 48 sec

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Which object has the most kinetic energy? a bus a car a plane a bicycle
loris [4]

Answer:I would guess a plane

Assuming they all Thad the same velocity....

4 0
3 years ago
The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?
Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours.  So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes.  So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

5 0
3 years ago
The length of the track itself is 500m. The vehicles (each of mass 15kg) are dragged
gogolik [260]

Answer:

W = 100000 J = 100 KJ

Explanation:

Here we will use the most basic and general formula of work, which is as follows:

W = Fd

where,

W = Work Done = ?

F = Force Required = 200 N

d = Length of Track = 500 m

Therefore,

W = (200\ N)(500\ m)\\

<u>W = 100000 J = 100 KJ</u>

5 0
3 years ago
A bird uses 10 N of force to pull a worm out of the ground a distance of 3 inches. How much work did the bird do?
pochemuha

Answer:

The work done by the bird is 0.762 J

Explanation:

Given;

force applied by the bird, f = 10 N

distance the bird moved the worm, d = 3 inches = 0.0762 m

The work done by the bird is given by;

W = F x d

where;

W is the work done by the bird

d is the distance the bird moved the load

Substitute the given values and estimate the work done by the bird;

W = 10 x 0.0762

W = 0.762 J

Therefore, the work done by the bird is 0.762 J

6 0
3 years ago
How can I become a good science student ?​
iris [78.8K]

Answer:

Study hard , focus on your studies and alyways ask questions .

7 0
2 years ago
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