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storchak [24]
3 years ago
11

Could I get some help with this

Physics
1 answer:
nika2105 [10]3 years ago
4 0

please give me brainlest!!

the answer is A.

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A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\
zloy xaker [14]

Answer:

  t = 1.77 s

Explanation:

The equation of a traveling wave is

       y = A sin [2π (x /λ -t /T)]

where A is the oscillation amplitude, λ the wavelength and T the period

the speed of the wave is constant and is given by

      v = λ f

Where the frequency and period are related

     f = 1 / T

we substitute

      v = λ / T

let's develop the initial equation

    y = A sin [(2π / λ) x - (2π / T) t +Ф]

where Ф is a phase constant given by the initial conditions

the equation given in the problem is

    y = 5.26 sin (1.65 x - 4.64 t + 1.33)

if we compare the terms of the two equations

 

         2π /λ = 1.65

          λ = 2π / 1.65

          λ = 3.81 m

         2π / T = 4.64

          T = 2π / 4.64

          T = 1.35 s

we seek the speed of the wave

           v = 3.81 / 1.35

           v = 2.82 m / s

           

Since this speed is constant, we use the uniformly moving ratios

          v = d / t

           t = d / v

           t = 5 / 2.82

           t = 1.77 s

3 0
3 years ago
A pitcher can throw a fastball that reaches home plate at 95 mph. What is this speed in m/s
Taya2010 [7]
42.47 meters per second
5 0
3 years ago
Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very
timofeeve [1]

Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ v=\frac{502.1}{\sqrt{3} }

      =289.9 \ m/s

The time will be:

⇒ t=\frac{d}{v}

      =\frac{2\times 6}{289.9}

      =\frac{12}{289.9}

      =0.041 \ sec

hence,

⇒ N=\frac{1}{t}

        =\frac{1}{0.041}

        =24.39 \ per \ sec

4 0
3 years ago
Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp
slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

Q=Av

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

Q=1.20 m^3/s

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2

So we can re-arrange the equation to find the speed of the water:

v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

Q_1 = Q_2\\A_1 v_1 = A_2 v_2

where we have

A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s

and where A_2 is the cross-sectional area of the pipe at the second point.

Solving for A2,

A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2

And finally we can find the radius of the pipe at that point:

A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

6 0
3 years ago
A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant
vovangra [49]

Answer:

emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.

Direction is counter clockwise.

Explanation:

See attached pictures.

6 0
3 years ago
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