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balandron [24]
3 years ago
12

Can someone help please

Physics
1 answer:
Katena32 [7]3 years ago
4 0
I’m pretty sure you would measure from crest to crest so A and B
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A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly
Mnenie [13.5K]

Answer:

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The acceleration of the tortoise is 0.28 m/s²

Explanation:

The equations that describe the position and velocity of the hare and the tortoise are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

To find the acceleration of the hare once it begins to slow down, we have to find how much time the hare traveled during the deceleration and what was its initial speed.

First, the hare moves with constant acceleration for 4.7 s. Then, its velocity at  t = 4.7 s will be:

v = v0 + a · t    (v0 = 0 because the hare starts form rest)

v = a · t = 0.9 m/s² · 4.7 s = <u>4.2 m/s</u>

<u />

The distance traveled by the hare while accelerating can be calculated using the equation of the position:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0 and v0 = 0)

x = 1/2 · a · t² = 1/2 · 0.9 m/s² · (4.7)² = <u>9.9 m</u>

<u />

Then, the hare runs at a constant speed of 4.2 m/s for 11.7 s. The distance traveled at constant speed will be:

x =  v · t

x = 4.2 m/s · 11.7 s = <u>49.1 m</u>

<u />

Then, the distance traveled by the hare while slowing down was:

Distance traveled while slowing down = 72 m - 49.1 m - 9.9 m = 13 m

Let´s find how much time it took the hare to come to stop, so we can calculate the acceleration. We know that when the position is 13 m, the velocity is 0.

v = v0 + a · t

0 = 4.2 m/s + a · t

-4.2 m/s / t = a

Replacing in the equation of the position:

x = v0 · t + 1/2 · a · t²      (considering x0 as the point at which the hare started to slow down)

13 m = 4.2 m/s · t - 1/2 · 4.2 m/s / t · t²

13 m = 4.2 m/s · t - 2.1 m/s · t

13 m = 2.1 m/s · t

t = 13 m / 2.1 m/s

t = 6.2 s

Then, the acceleration of the hare while slowing down will be:

-v0/t = a

-4.2 m/s / 6.2 s = a

a = -0.68 m/s²

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The hare traveled 72 m in (6.2 s + 11.7 s + 4.7 s) 22.6 s. The tortoise reaches the final position of the hare at the same time, so, using the equation of the position we can calculate the acceleration of the tortoise:

x = x0 + v0 · t + 1/2 · a · t²     (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

72 m = 1/2 · a · (22.6 s)²

144 m / (22.6 s)² = a

a = 0.28 m/s²

The acceleration of the tortoise is 0.28 m/s²

6 0
3 years ago
A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of i
kirza4 [7]

Answer:

2.5 x 10⁷ J

Explanation:

F = thrust of the engine = 2.3 x 10⁵ N

d = distance traveled = 87  m

Work done by the engine is given as

W = F d =  (2.3 x 10⁵) (87) = 200.1 x 10⁵ J

W' = Net work done

W'' = work done by catapult

KE₀ = initial kinetic energy = 0 J

KE = final kinetic energy = 4.5 x 10⁷ J

Net work done is given as

W' = KE - KE₀

W' = 4.5 x 10⁷ J

We know that

W' = W + W''

4.5 x 10⁷ = 2.001 x 10⁷ + W''

W'' = 2.5 x 10⁷ J

8 0
3 years ago
What is a model drawing of 2HCI?
klasskru [66]
This is the answer just double it

7 0
3 years ago
A 2.5 kg mass starts from rest at point A and moves along the x-axis subject to the potential energy shine in the figure below
expeople1 [14]
It’s definitely b Yk Why um I don’t really know butttt gut instinct
3 0
4 years ago
Increasing the concentration of greenhouse gases in Earth's atmosphere decreases the transparency of the atmosphere to infrared
Artist 52 [7]

Answer:

Decrease the transparency of the atmosphere

Explanation:

Different gases have characteristic absorption bands, so-called greenhouse gases have very high bands in the infrared region and small bands in the region of visible light.

   When the radiation  arrives to pass the greenhouse gases (CO²) and affects the earth's surface after several processes, the earth emits radiation in the infrared region, but this is absorbed by the gases, creating a net heating, since that the transparency of the atmosphere is  DECREASE for infrared radiation

Reviewing the correct answers is: Decrease the transparency of the atmosphere

5 0
3 years ago
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