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devlian [24]
3 years ago
5

PLEASE HELPPPP!!!!

Physics
2 answers:
ivanzaharov [21]3 years ago
5 0

Answer:

The right option is higher pitch due to change to a higher frequency

Explanation:

Pitch can be defined as the characteristics of a note which enables us to differentiate a high note from a low one.

Frequency: this is the number of oscillation per seconds expressed in Hertz (Hz)

Pitch of a note depends on frequency of a sound wave

I.e as the frequency become high, The higher will be the pitch.

From the question above, when the sound source is moving towards the observer,

Let the frequency of the sound source = f

Let the apparent frequency (frequency which the observer will hear) = f¹

Let the velocity of the source sound = Vₓ

Let the velocity of the observer = V

∴ The apparent frequency (f¹) when the source sound is moving towards the observer = {V/(V-Vₓ)}f

f¹ = {V/(V-Vₓ)}f

since (V-Vₓ ) is less than V, therefore f¹ will be increased.

If f¹ is increased, the pitch is also increased.

The right option is higher pitch due to change to a higher frequency

Tresset [83]3 years ago
3 0

you will hear a higher pitch due to a higher frequency.

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Answer: Imma say balanced

Explanation:

Because if u pull one side of a skate board at 10n and your friend pulls the other side of the skate board at 10n it will be balanced because it doesn't move.

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A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots
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(a). The car's average velocity between t = 1.0s to t = 1.5s will be - 1\;m/s

(b). The car's acceleration at t = 1.5s will be - 0.4\;m/s^{2}

(c). Car's speed is increasing with time.

We have a a remote controlled toy car that starts from rest and begins to accelerate in a straight line.

We have to determine -

  • The car's average velocity (in m/s) in the interval between -

        t = 1.0 s  to  t = 1.5 s.

  • The car's acceleration at t = 1.5 s.
  • Determining whether car's speed increasing or decreasing with time.

<h3>What is Acceleration?</h3>

The rate of change of velocity with respect to time is called Acceleration. Mathematically -

$a=\frac{dv}{dt}

According to the question, we have the following data for the Car -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m

PART - A

The car's average velocity between t = 1.0s to t = 1.5s will be -

$v_{avg} = \frac{0.9-0.4}{1.5-1}= 1 m/s

PART - B

Velocity at t = 1.5 s will be -

$v(1.5)=\frac{0.9}{1.5}= 0.6\;m/s

The car's acceleration at t = 1.5s will be -

$a(1.5) = \frac{v}{t} = \frac{0.6}{1.5} = 0.4\;m/s^{2}

PART - C

Since, the acceleration of the car is positive, this means that the car is accelerating in the forward direction. Hence, its speed is increasing with time.

[ The following data was missing in your answer. The complete question would include this data also -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m ]

To solve more questions on Kinematics, visit the link below-

brainly.com/question/17272824

#SPJ2

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Answer:

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During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket h
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Answer:

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Explanation:

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The tangential component is:

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The radial component is:

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ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

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