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IRISSAK [1]
3 years ago
11

A device called oacillatwr is used to swnd waves along a streatched string to send waves along a streatched string. The string i

s 20 cm long, and four complete waves fit along its length when the oscillation vibrates 30 times per second the waves on the string.. What is its frequency and speed and wavelength
Physics
1 answer:
vitfil [10]3 years ago
3 0

Answer:

Frequency = 2m/s1m=2 waves/s, or 2 Hz

A wave is traveling at a speed of 2 m/s and has a frequency of 2 Hz.

Explanation:

wd Calculating Wave Frequency or Wavelength from Wave Speed

The equation for wave speed:

Frequency = SpeedWavelength or Wavelength = SpeedFrequency

Therefore, if you know the speed of a wave and either the wavelength or wave frequency, you can calculate the missing value.

For example, suppose that a wave is traveling at a speed of 2 meters per second and has a wavelength of 1 meter. Then the frequency of the wave is:

- Frequency = 2m/s1m=2 waves/s, or 2 Hz

A wave is traveling at a speed of 2 m/s and has a frequency of 2 Hz. What is its wavelength?

- Substitute these values into the equation for wavelength:

- Wavelength = 2m/s2waves/s=1 m

See more: https://www.ck12.org/physics/wave-speed/lesson/Wave-Speed-MS-PS/

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Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is r
katen-ka-za [31]

Answer:

<em>171.5m</em>

Explanation:

The velocity of sound in water = 343m/s

Time taken = 1.00secs

using the formula to calculate the distance

2x = vt

x is the distance

v is the speed of sound

t is the time

x = vt/2

x = 343(1)/2

x = 171.5m

<em>hence their separation 1.00 s after the second object is released is 171.5m</em>

4 0
3 years ago
You are an engineer helping to design a roller coaster that carries passengers down a steep track and around a vertical loop. Th
vova2212 [387]

Answer:

h >5/2r

Explanation:

This problem involves the application of the concepts of force and the work-energy theorem.

The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.

Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)

So from newton's second law of motion,

W – N = mv²/r

N = normal force = 0

W = mg

mg = ma = mv²/r

mg = mv²/r

v²= rg

v = √(rg)

The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop

So

ΔPE = ΔKE = 1/2mv²

The height at the roller coaster starts is usually higher than the top of the loop by design. So

ΔPE =mgh - mg×2r = mg(h – 2r)

2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.

In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.

So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.

ΔPE > ΔKE

mg(h–2r) > 1/2mv²

g(h–2r) > 1/2(√(rg))²

g(h–2r) > 1/2×rg

h–2r > 1/2×r

h > 2r + 1/2r

h > 5/2r

5 0
3 years ago
Read 2 more answers
What is the speed of a wave on a string with a wavelength of 1.75 m and a frequency of 2.0 Hz
deff fn [24]

Answer:

V=3.5 m/s

Explanation:

V=(F)(W)

V=(2)(1.75)

V= 3.5 m/s

7 0
3 years ago
imagine you are going on a rid in a spacecraft next to earth. Your trip takes one whole year. Describe earth's tilt in the north
mylen [45]
You will have to fly around the whole earth to get to your landing station
8 0
3 years ago
Which of the following have the same density. number 26
marusya05 [52]

1 cubic cm is the same as 1 mL, so the answer would be C.

8 0
3 years ago
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