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timofeeve [1]
2 years ago
8

Suppose that an owner of the same dog breed has also taken some measurements. They notice that the surface area of the dog has i

ncreased by a factor of 3 over a period of four years. Determine the change in the dog's relative surface area in this time period.
Physics
1 answer:
Mnenie [13.5K]2 years ago
7 0

Answer:

Surface area of the dog is changes from A to 3A

Explanation:

It is given that surface area of dog is increased by factor 3 in a period of 4 year

We have to find the change in dog's relative surface area in the given time period

Let initially the surface area is A

As the surface area is increased by a factor of 3

So surface area after 4 year = 3×A = 3A

So surface area of the dog is changes from A to 3A

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Answer:The higher up an object is the greater its gravitational potential energy. The larger the distance something falls through the greater the amount of GPE the object loses as it falls. As most of this GPE gets changed into kinetic energy, the higher up the object starts from the faster it will be falling when it hits the ground.  So a change in gravitational potential energy depends on the height an object moves through.

Explanation: Lifting an apple up 1 metre is easier work than lifting an apple tree the same height. This is because a tree has more mass, so it needs to be given more gravitational potential energy to reach the same height.

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A cat falls from a table of height 1.3 m. What is the impact speed of the cat?
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Use this formula
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In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th
qaws [65]

Answer:

The distance between both cars is 990 m

Explanation:

The equations for the position and the velocity of an object moving in a straight line are as follows:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a * t

where:

x = position of the car at time "t"

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = velocity

First let´s find how much time it takes the driver to come to stop (v = 0).  We will consider the origin of the reference system as the point at which the driver realizes she must stop. Then x0 = 0

With the equation of velocity, we can obtain the acceleration and replace it in the equation of position, knowing that the position will be 250 m at that time.

v = v0 + a*t

v-v0 / t = a

0 m/s - 71.0 m/s / t =a

-71.0 m/s / t = a

Replacing in the equation for position:

x = v0* t +1/2 * a * t²

250 m = 71.0 m/s * t + 1/2 *(-71.0 m/s / t) * t²

250 m = 71.0 m/s * t - 1/2 * 71.0 m/s * t

250m = 1/2 * 71.0m/s *t

<u>t = 2 * 250 m / 71.0 m/s = 7.04 s</u>

It takes the driver 7.04 s to stop.

Then, we can calculate how much time it took the driver to reach her previous speed. The procedure is the same as before:

v = v0 + a*t

v-v0 / t = a      now v0 = 0 and v = 71.0 m/s

(71.0 m/s - 0 m/s) / t = a

71.0 m/s / t =a

Replacing in the position equation:

x = v0* t +1/2 * a * t²      

390 m = 0 m/s * t + 1/2 * 71.0 m/s / t * t²       (In this case, the initial position is in the pit, then x0 = 0 because it took 390 m from the pit to reach the initial speed).

390m * 2 / 71.0 m/s = t

<u>t = 11.0 s</u>

In total, it took the driver 11.0s + 5.00 s + 7.04 s = 23.0 s to stop and to reach the initial speed again.

In that time, the Mercedes traveled the following distance:

x = v * t = 71.0 m/s * 23.0 s = 1.63 x 10³ m

The Thunderbird traveled in that time 390 m + 250 m = 640 m.

The distance between the two will be then:

<u>distance between both cars = 1.63 x 10³ m - 640 m = 990 m.  </u>

3 0
2 years ago
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