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sp2606 [1]
3 years ago
7

A thin, 75.0 cm wire has a mass of 16.5 g. One end is tied to a nail, and the other is attached to a screw that can be adjusted

to vary the tension in the wire. (a) To what tension in Newtons must you adjust the screw to that a transverse wave of wavelength 3.33 cm makes 875 vibrations per second? (b) How fast would this wave travel?
Physics
1 answer:
natka813 [3]3 years ago
8 0

Answer:

(a) 29.14 m/s

(b) 18.7 N

Explanation:

given information:

L = 75 cm = 0.75 m

mass, m = 16.5 g = 1.65 x 10⁻² kg

wavelength, λ = 3.33 cm = 0.0333 m

frequency, f = 875

first we calculate the speed

v = λf

  = 0.0333 x 875

  = 29.14 m/s

v = √F/μ

where

v = speed

F = tension

μ = linear density

μ = m/L, m is mass, and L is the length

thus,

v² = F/(m/L)

F = v²m/L

  = (29.14)²(1.65 x 10⁻² )/(0.75)

  = 18.7 N

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the answer is B! it would continue to expand.....just took the test XD

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alex41 [277]

The sum of the series allows to find the result for the total distance that the ball bounces is:

            total distance = 59.52 in

A series is a set of things or numbers related by a specific operation.

They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.

Let's build a table to observe the sequence.

drop   height    rebound

   1        30          15

   2       15            7.5

   3        7.5         3.75

If we call the first term y₀  

The first bounce can be found.

                             y₁ = \frac{y_o}{2}

The second bounces.

                             y_2 = \frac{y_1}{2}  \\y_2 = \frac{y_o}{4}

The third bounce.

                             y_3  = \frac{y_2}{2}  \\y_3 = \frac{y_0}{ 8}

By observing this table we can construct a series of the form

 

      Total distance = y_o \ ( 1 + \frac{1}{2} + \frac{1}{4}+  \frac{1}{8} + ... +\frac{1}{2n} )

The sum of the serie has a result of

        sum  = 127/64 = 1,984

Let's calculate

     distance total  = 30 1,984

     Distance total = 59.52 cm

In conclusion, using the sum of the series we can find the result for the total distance that the ball bounces is:

            total distance = 59.52 in

Learn more here: brainly.com/question/8879163

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ate around its central axis. A rope wrapped around the drum of radius 1.24 m exerts a force of 4.56 N to the right on the cylind
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Answer:

Magnitude the net torque about its axis of rotation is 1.3338 Nm

Explanation:

The radius of the wrapped rope around the drum, r = 1.24 m

Force applied to the right side of the drum, F = 4.56 N

The radius of the rope wrapped around the core, r' = 0.57 m

Force on the cylinder in the downward direction, F' = 7.58 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

\tau = F\times r\tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'\tau_{net} = - 4.56\times 1.24 + 7.58\times 0.57 \\\\= - 1.3338\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

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Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At w
valentina_108 [34]

Answer:before throwing and after catching the ball

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Main Answer:

Given  speed of spaceship v = 0.8c = 0.8 * 3 x 10^8 m/s

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we know that 1 light year  = 9.461 x 10^15 m

Distance need to be travelled d = 4.3 x 9.461 x 10^15

d = 40.6823 x 10^15 m

Time taken for the trip would elapse on a clock on board the spaceship

t = distance/ velocity

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t = 16.95 x 10^7 sec

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Explanation:

What is light year?

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