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1 year ago

What is the displacement of the iguana between 0s and 5s?

Physics

2 answers:

butalik [34]1 year ago

5 0

Answer:

18 meters

Explanation:

The displacement is the area under the graph.

The equation of the line from x=0 to x=3 is y = 8/3 x − 2. The x-intercept is (3/4, 0).

The area of the negative triangle is:

½ (3/4) (-2) = -0.75

The area of the positive trapezoid is:

½ ((5 − 3) + (5 − 3/4)) (6) = 18.75

The total area is 18.75 + -0.75 = 18.

Vinil7 [7]1 year ago

5 0

Answer: the displacement is 8m and the distance traveled is also 8 meters

Explanation:

It travels from -2 to 6, a difference of positive 8, and since that was its initial starting point, displacement is 8m, distance traveled also happens to be 8m

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Rank the following situations according to the magnitude of the impulse of the net force, from largest value to smallest value.

wolverine [178]

Answer:

I and II

III and IV

Explanation:

The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.

Taking always east as positive direction, and labelling

u the initial velocity

v the final velocity

m = 1000 kg the mass (which is always equal)

We find:

(i)

u = 25 m/s

v = 0

$|I|=m(v-u)=(1000)(0-25)=25,000 Ns$

(II)

u = 25 m/s

v = 0

$|I|=m(v-u)=(1000)(0-25)=25,000 Ns$

(III)

In this case,

F = 2000 N is the force

$\Delta t = 10 s$ is the time

So the magnitude of the impulse is

$|I| =F\Delta t = (2000N)(10)=20,000 Ns$

(IV)

F = 2000 N is the force

$\Delta t = 10 s$ is the time

So the magnitude of the impulse is

$|I| =F\Delta t = (2000N)(10)=20,000 Ns$

(V)

u = 25 m/s

v = -25 m/s

$|I|=m(v-u)=(1000)(-25-25)=50,000 Ns$

So the ranking from largest to smallest is:

I and II

III and IV

5 0

2 years ago

A physics student with too much free time drops a watermelon from a roof of a building, hears the sound of the watermelon going

tatiyna

Answer:

28.6260196842 m

Explanation:

Let h be the height of the building

t = Time taken by the watermelon to fall to the ground

Time taken to hear the sound is 2.5 seconds

Time taken by the sound to travel the height of the cliff = 2.5-t

Speed of sound in air = 340 m/s

For the watermelon falling

$s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow h=\frac{1}{2}\times 9.81\times t^2$

For the sound

Distance = Speed × Time

$\text{Distance}=340\times (2.5-t)$

Here, distance traveled by the stone and sound is equal

$\frac{1}{2}\times 9.81\times t^2=340\times (2.5-t)\\\Rightarrow 4.905t^2=340\times (2.5-t)\\\Rightarrow t^2=\frac{340}{4.905}(2.5-t)\\\Rightarrow t^2+69.3170234455t-173.292558614=0$

$t=\frac{-69.31702\dots +\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1},\:t=\frac{-69.31702\dots -\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1}\\\Rightarrow t=2.4158\ s\ or\ -71\ seconds$