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Vedmedyk [2.9K]
3 years ago
10

A sinusoidal transverse wave travels along a long, stretched, string. the amplitude of this wave is 0.0885 m, it's frequency is

4.31 hz, and its wavelength is 1.21m.
(a) What is the transverse distance between a maximum and a minimum of the wave?
(b) How much time is required for 71.7 cycles of the wave to pass a stationary observer?
(c) Viewing the whole wave at any instant, how many cycles are there in a 30.7-m length of string?
Physics
1 answer:
timofeeve [1]3 years ago
6 0

Answer:

(a) 0.177 m

(b) 16.491 s

(c) 25 cycles

Explanation:

(a)

Distance between the maximum and the minimum of the  wave = 2A ............ Equation 1

Where A = amplitude of the wave.

Given: A = 0.0885 m,

Distance between the maximum and the minimum of the wave = (2×0.0885) m

Distance between the maximum and the minimum of the wave = 0.177 m.

(b)

T = 1/f ...................... Equation 2.

Where T = period, f = frequency.

Given: f = 4.31 Hz

T = 1/4.31

T = 0.23 s.

If 1 cycle pass through the stationary observer for 0.23 s.

Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.

= 16.491 s.

(c)

If  1.21 m contains  1 cycle,

Then, 30.7 m will contain (30.7×1)/1.21

= 25.37 cycles

Approximately 25 cycles.

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when a circular plate of metal is heated in an oven, its radius increases at .03 cm/min, at what rate is the area increasing whe
Alinara [238K]

Answer:

Rate of change of area will be 9.796cm^2/min

Explanation:

We have given rate of change of radius \frac{dr}{dt}=0.03cm/min

Radius of the circular plate r = 52 cm

Area is given by A=\pi r^2

So \frac{dA}{dt}=2\pi r\frac{dr}{dt}

Puting the value of r and \frac{dr}{dt}

\frac{dA}{dt}=2\times 3.14\times 52\times 0.03=9.796cm^2/min

So rate of change of area will be 9.796cm^2/min

6 0
3 years ago
The position of a car at time t is given by the function p(t)=t2 2t−4. What is the velocity when p(t)=11? assume t≥0
AysviL [449]

The velocity when function p(t)=11 is 8 .

According to the question

The position of a car at time t  represented by function :

p(t)=t^{2} +2t-4

Now,

When  function p(t) = 11 , t will be

p(t)=t^{2} +2t-4

11 = t²+2t-4

0 = t² + 2t - 15

or

t² +2t-15 = 0

t² +(5-3)t-15 = 0

t² +5t-3t-15 = 0

t(t+5)-3(t+5) = 0

(t-3)(t+5) = 0  

t = 3 , -5  

as t cannot be -ve as given ( t≥0)

so,

t = 3

Now,

the velocity when p(t)=11

As we know velocity = \frac{position}{time}

therefore to get the value of velocity from  function p(t)

we have to differentiate the function with respect to time

\frac{d(p(t))}{dt} =\frac{d}{dt} (t^{2} +2t-4)

v(t) = 2t + 2  

where v(t) = velocity at that time

as t = 3 for  p(t)=11  

so ,

v(t) = 2t + 2  

v(t) = 2*3 + 2  

v(t) = 8

Hence, the velocity when function p(t)=11 is 8 .

To know  more about function here:

brainly.com/question/12431044

#SPJ4

4 0
2 years ago
Acceleration increases over time once a force is applied to the object. Determine the acceleration at 3.5 sec? A) 3 m/s2 B) 6 m/
Novay_Z [31]

Answer: Acceleration does not increase over time once a force is applied to the object. It depends on the force and the object's mass. If those don't change, then the acceleration is constant.

Explanation:. F = M • a

We don't know either of those numbers, so we can't answer the question.

4 0
3 years ago
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melisa1 [442]

Answer:

6

Explanation:

3 0
3 years ago
The graph above shows the position and time calculate the velocity of the particle from T=0s to T=4s?
zzz [600]

Answer:

The velocity of the particle from T = 0 s to T = 4 s is;

0.5 m/s

Explanation:

The given parameters from the graph are;

The initial displacement (covered) at time, t₁ = 0 s is x₁ = 1 m

The displacement covered at time, t₂ = 4 s is x₂ = 3 m

The graph of distance to time, from time t = 0 to time t = 4 is a straight line graph, with the velocity given by the rate of change of the displacement to the time which is dx/dt which is also the slope of the graph given as follows;

The \ slope \ of \ the \ displacement \ time \ graph, \ m =velocity, \ v= \dfrac{x_2 - x_1}{t_2 - t_1}

Therefore, \ the \ velocity \ of \ the \ particle \ v  = \dfrac{3 \  m - 1 \ m}{4 \ s - 0 \ s}  = \dfrac{2 \ m}{4 \  s} = \dfrac{1}{2} \ m/s

The velocity of the particle from t = 0 s to t  = 4 s = 1/2 m/s = 0.5 m/s.

3 0
3 years ago
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