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denpristay [2]
2 years ago
9

A sound wave traveling at 340 m/s is generated by a 480 Hz tuning fork.

Physics
1 answer:
Shtirlitz [24]2 years ago
8 0

Answer:

Wavelength = 0.7083 meters

Explanation:

Given the following data;

Speed of wave = 340 m/s

Frequency = 480 Hz

To find how long is the sound wave, we would determine its wavelength;

Mathematically, the wavelength of a waveform is given by the formula;

Wavelength = velocity/frequency

Wavelength = 340/480

Wavelength = 0.7083 meters

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2) The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. How long doe
ValentinkaMS [17]

Answer: 2.04 s

Explanation:

Let the initial velocity be v, Angle of projectile be

Then the horizontal component = v cos θ = 16 m/s

Vertical component of velocity = v sin θ = 20 m/s

Time taken to reach the highest point is half the time taken for total flight.

Time for total flight,

t = \frac{2vsin \theta}{g}

t'=\frac{vsin \theta}{g} = \frac {20 m/s}{9.8 m/s^2} = 2.04 s

Thus, the football takes 2.04 s to rise to the highest point of its trajectory.

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3 years ago
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Answer:

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Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre
castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:  

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)  

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619  

=

t

+

C

Here C =constant of integration.   

x

=

0  at  t

=

0

, we get:  C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619  =

t

−

4.7619

.

.

.

.

.

.

.

.

.

(

2

)

Here

x  is measured in miles and  t  in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),    

     to get:

      = −

4.7619  

      =  

1

−

4.7619

      = −

3.7619

  or  x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in   m

i

l

/

differentiate  

     equation (1) with respect to time.

     we have to eliminate x from the equation (1) using equation (2).  

     Eliminating x we get:

     v

=

7.5×

     Now differentiating above equation w.r.t time we get:

      a

=

d

v/

d

t

       =

−

0.675

/

      At  

      t

=

0

      the joggers acceleration is :

       a

=

−

0.675

m

i

l

/

        =

−

4.34

×

f

t

/  

(c)  required time for the jogger to run 6 miles is obtained by setting  

        x

=

6  in equation (2).  We get:

        −

4.7619

(

1

−

(

0.04

×

6  )

)^

7

/

10=

t

−

4.7619

         or

         t

=

0.832

h

r

s

6 0
3 years ago
This is a velocity versus time graph of a car starting from rest. If the area under the line is 10 meters, what is the correspon
adelina 88 [10]

So the area under a velocity time graph is distance or displacement, if you have done calculus yet you will understand that if you take the integral of a velocity function then you end up with displacement. Thats for later understanding however.

So this appears to be a right triangle so we can find the area of a triangle as:

0.5bh = A

Since our area is 10 meters lets alter our formula a bit to fit the situation:

Our base here is time and our height is velocity so:

0.5tv = Δx

So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s

So we have v, and Δx so lets isolate for time by dividing by v and 0.5

t = Δx / 0.5v

Now lets plug all that in:

t = 10 / 0.5(2)

t = 10 seconds

Hope this helped!

8 0
3 years ago
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