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Crazy boy [7]
2 years ago
14

Looking at the bottom standing wave in the picture, how many complete waves are present?

Physics
1 answer:
Shtirlitz [24]2 years ago
4 0

Answer:

the last one

Explanation:

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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
3 years ago
Similarities and differences between<br> distance Time graph and velocity time<br> graph?
egoroff_w [7]

Answer:

Similarity: >>Time is independent variable and such is on the x-axis. ... >>Distance time graph tells you how much distance you have travelled, while velocity time graph tells you your acceleration.  The difference between them is that the velocity-time graph reveals the speed of an object (and whether it is slowing down or speeding up), while the position-time graph describes the motion of an object over a period of time.

Explanation:

4 0
3 years ago
What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?​
givi [52]

Answer:

1.2 x 10^-12

Explanation:

3/2.5 x 10^-6/10^6

1.2 x 10^-6 x 10^-6

1.2 x 10^-12

6 0
3 years ago
A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.
melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

V_{f}^{2} = V_{0}^{2} + 2ad

<u>Where</u>:

V_{f}: is the final speed = 8.89 m/s

V_{0}: is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m    

a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}

Therefore, the magnitude of her acceleration is 3.09 m/s².              

b) The component of her acceleration that is parallel to the ground is given by:

a_{x} = a*cos(\theta)

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

7 0
3 years ago
I didn't understand ​
julsineya [31]
Huh? The answer is A Shadow
4 0
3 years ago
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