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3 years ago

How could the Doppler Effect have assisted you with this investigation?

1 answer:

4 0

If you started moving either away or twords the object of noise you could use a noise dedicator to see if the waves change in pitch or volume

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Find the final velocity if the initial velocity of 8 m/s with an acceleration of 7 m/s2 over a 3 second interval?

LenKa [72]

I don't know about it your answer will give another people

5 0

3 years ago

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A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to

dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to $T=2\pi \sqrt{\frac{m}{K}}$ , here m is mass and K is spring constant

So $1.41=2\times 3.14\sqrt{\frac{0.683}{K}}$

$0.2245=\sqrt{\frac{0.683}{K}}$

Squaring both side

$0.0504=\frac{0.4664}{K}$

$K=9.255N/m$

So spring constant of the spring will be equal to 9.255 N /m

7 0

4 years ago

During baseball practice, a batter hits a high flyball, and then runs in a straight line and catches it. which has the greater a

Jlenok [28]

They have the same velocity because their displacements (shortest line from point A to point B, which is a straight line) are the same and they meet at the same time.

6 0

3 years ago

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What is the Kinetic Energy of a 120 kg object that is moving with a velocity of 15 m/s

gizmo_the_mogwai [7]

Answer:

Kinetic Energy:120 x 15=1800

Explanation:

8 0

3 years ago

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The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 440g . The vibra

ANEK [815]

Answer:

Tension, T = 2038.09 N

Explanation:

Given that,

Frequency of the lowest note on a grand piano, f = 27.5 Hz

Length of the string, l = 2 m

Mass of the string, m = 440 g = 0.44 kg

Length of the vibrating section of the string is, L = 1.75 m

The frequency of the vibrating string in terms of tension is given by :

$f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$

$\mu=\dfrac{m}{l}$

$\mu=\dfrac{0.44}{2}=0.22\ kg/m$

$T=4L^2f\mu$

$T=4\times (1.75)^2\times (27.5)^2 \times 0.22$

T = 2038.09 N

So, the tension in the string is 2038.09 N. Hence, this is the required solution.

6 0

4 years ago