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Rama09 [41]
2 years ago
3

A transverse wave with a frequency of 125 Hz is traveling in the x-direction. An illustration of a transverse wave with the dist

ance from the high point to the low point labeled 5 centimeters and the vertical distance from the high point of the wave to the low point of the wave labeled 12 m. How long will it take this wave to travel 3000 m in the x-direction?
Physics
2 answers:
Vsevolod [243]2 years ago
6 0

Answer:

t = 240 s

Explanation:

Given:-

- The frequency, f = 125 Hz

- distance from the high point to the low point labeled 5 centimeters

- distance from the high point to the low point labeled 5 centimeters

Find:-

How long will it take this wave to travel 3000 m in the x-direction?

Solution:-

- Distance from the high point to the low point labeled = 5 centimeters denotes half of a cycle of wave or 1/2*wavelength (λ).

- The wavelength (λ) of this wave is:

                               (λ) = 2*5 = 10 cm

- The relationship between speed (v) of wave and its wavelength (λ) and frequency (f) is given by:

                               v = f*λ

                               v = 125*(0.1)

                               v = 12.5 m/s

- The time (t) taken by the wave to travel a distance d = 3000 m in x-direction can be evaluated by the formula:

                               t = d / v

                               t = ( 3000 ) / 12.5

                              t = 240 s

ELEN [110]2 years ago
3 0

Answer:

2.4 seconds

Explanation:

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Answer:

Explanation:

The formula for gravitational potential energy is

Ep = m · g · h   Assuming that the acceleration is g = 10m/s²

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2 years ago
A dog of mass 10 kg sits on a skateboard of mass 2 kg that is initially traveling south at 2 m/s. The dog jumps off with a veloc
Tasya [4]

Answer:

17 m/s south

Explanation:

m_1 = Mass of dog = 10 kg

m_2 = Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

u_1 = Velocity of dog = 1 m/s

u_2 = Velocity of skateboard

In this system the linear momentum is conserved

(m_1+m_2)v+m_1u_1+m_2u_2=0\\\Rightarrow u_2=-\dfrac{(m_1+m_2)v+m_1u_1}{m_2}\\\Rightarrow u_2=-\dfrac{(10+2)2+10\times 1}{2}\\\Rightarrow u_2=-17\ m/s

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3 0
2 years ago
You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As so
grigory [225]

Answer:

Explanation:

Time duration during which acceleration exists in  bicycle =

23 / 12 = 1.91 s

Time duration during which acceleration exists in car

= 47 / 8 = 5.875 s

Distance covered by bicycle during acceleration ( t = 1.91 s )

= 1/2 x 12 x (1.91)²

= 21.88 mi

Distance covered by car during this time ( t = 1.91 s )

= 1/2 x 8 x (1.91)²

7.64 mi ,

velocity of car after 1.91 s

= 8 x 1.91 = 15.28 mi/h

Let after time 1.91 , time taken by them to meet each other be t

Total distance covered by cycle = total distance covered by car

21.88 + 23 t = 7.64 + 15.28t + 4 t²

21.88 = 7.64 - 7.72t +4 t²

4 t² -7.72 t -14.24 = 0

t = 2.83 s

Total time taken

= 2.83 + 1.91

= 4.74 s

So after 4.74 s they will meet each other.

b ) Maximum distance occurs when velocity of both of them becomes equal .

Velocity after 1.91 s of bicycle

12 x 1.91 = 23 mi/h

Velocity after 1.91 s of car

8 x 1.91 = 15.28 mi/h . Let after time t , the velocity of car becomes 23

15.28 + 8t = 23

t = .965 s

So after time .965 s , car has velocity equal to that of bicycle.

The bicycle will travel a distance of

= 21.88 + .965 x 23 = 44.075 mi

car will travel a distance of

7.64 + 15.28 x .965 + .5 x 8 x .965²

= 7.64 + 14.75 + 3.72

= 26.11 mi

Distance between car and bicycle

= 44.075 - 26.11 = 17.965 mi

= 17.965 x 1760

= 31618.4 ft.

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Answer:

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