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love history [14]
3 years ago
5

In a swimming meet, the swimmers swim a total of 8 laps of a 50-meter-long swimming pool. What is the distance traveled by a swi

mmer in this meet? What is swimmer’s displacement?
Physics
1 answer:
N76 [4]3 years ago
7 0

Answer:

The swimmer has a distance traveled of 800 meters.

The final displacement of the swimmer is 0 meters.

Explanation:

A lap is a round trip made by a swimmer in the pool, so that the distance traveled by swimmer is sixteen times the length of the swimming pool. That is:

s = \left(2\,\frac{travels}{lap} \right)\cdot \left(8\,laps \right)\cdot \left(50\,\frac{m}{travel}\right)

s = 800\,m

A swimmer has a distance traveled of 800 meters.

The displacement is the distance between swimmer and a reference point, let suppose that reference point is located at the beginning of the first lap. Hence, the final displacement of the swimmer is 0 meters.

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Tim and Rick both can run at speed Vr and walk at speed Vw, with Vr > Vw.
miss Akunina [59]

Answer:

Δt =  \frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw}

Explanation:

Hi there!

Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.

The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):

v = d/t

Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:

Vr = D/ (2 · Δt1)

For the other half of the trip the expression of velocity will be:

Vw = D/(2 · Δt2)

The total time traveled is the sum of both Δt:

Δt(total) = Δt1 + Δt2

Then, solving the first equation for Δt1:

Vr = D/ (2 · Δt1)

Δt1 = D/(2 · Vr)

In the same way for the second equation:

Δt2 = D/(2 · Vw)

Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)

Δt(total) = D/2 · (1/Vr + 1/Vw)

The time needed by Rick to complete the trip was:

Δt(total) = D/2 · (1/Vr + 1/Vw)

Now let´s calculate the time it took Tim to do the trip:

Tim walks half of the time, then his speed could be expressed as follows:

Vw = 2d1/Δt  Where d1 is the traveled distance.

Solving for d1:

Vw · Δt/2 = d1

He then ran half of the time:

Vr = 2d2/Δt

Solving for d2:

Vr · Δt/2 = d2

Since d1 + d2 = D, then:

Vw · Δt/2 +  Vr · Δt/2 = D

Solving for Δt:

Δt (Vw/2 + Vr/2) = D

Δt = D / (Vw/2 + Vr/2)

Δt = D/ ((Vw + Vr)/2)

Δt = 2D / (Vw + Vr)

The time needed by Tim to complete the trip was:

Δt = 2D / (Vw + Vr)

Let´s find the diference between the time done by Tim and the one done by Rick:

Δt(tim) - Δt(rick)

2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))

\frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw} = Δt

Let´s check the result. If Vr = Vw:

Δt = 2D/2Vr - D/2Vr - D/2Vr

Δt = D/Vr - D/Vr = 0

This makes sense because if both move with the same velocity all the time both will do the trip in the same time.

8 0
4 years ago
In order to increase the amount of work completed it is necessary to
navik [9.2K]
Do all you can in one big day that you have time off or work on one thing then work on the other at the same time

3 0
3 years ago
Read 2 more answers
Need help please!! I literally am dumb and can’t do this
natta225 [31]

Answer:

The answer should be A and B. I don't fully know so don't quote me on that! Haha.

Explanation:

6 0
4 years ago
A quantity that has only one magnitude
Len [333]

Answer:

scalar quantity

Explanation:

Vector quantities have two characteristics, a magnitude and a direction. Scalar quantities have only a magnitude.

3 0
4 years ago
A 5.00X10^5 kg rocket is accelerating straight up. Its engines produce 1.250X10^7 N of thrust, and air resistance is 4.50X10^6 N
katrin [286]
<span>6.20 m/s^2 The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be 9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N Add in the atmospheric drag and you get 4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N Now subtract that total drag from the thrust available. 1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So 3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2 Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2</span>
8 0
3 years ago
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