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2 years ago

A wave travels a distance of 60cm in 3s. The distance blw successive crests of thd wave is 4cm. What is the frequency?

1 answer:

6 0

Data:
ΔS = 60cm
Δt = 3s
Vm = ?

Vm = ΔS/Δt
$Vm = \frac{60}{3}$
$Vm = 20cm/s$

We have:

Vm = v
ΔS = λ
Δt = T

$v = \frac{\lambda}{T}$
$20 = \frac{\lambda}{4}$
$\lambda = 20*4$
$\lambda = 80cm$

Soon:

$v = \lambda*f$
$20 = 80*f$
$80f = 20$
$f = \frac{20}{80} simplify( \frac{\div20}{\div20} )= \frac{1}{4}$
$\boxed{f = 0,25Hz}$

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2 years ago

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Answer:

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2 years ago

A baseball player friend of yours wants to determine his pitching speed. you have him stand on a ledge and throw the ball horizo

zhenek [66]

Answer:

The pitching speed of your friend is 33.20 m/s

Explanation:

Lets explain how to solve the problem

Your friend throw the ball horizontally that means the vertical initial

component of velocity is zero ( $u_{y}=0$ ).

The ball is thrown from a height 4 meters above the ground.

The height $h=u_{y}t+\frac{1}{2}gt^{2}$

Remember: the height is negative value because its below the point of

thrown (initial position)

h = -4 m , $u_{y}=0$ and g = -9.8 m/s²(downward)

Substitute these values in the rule above

⇒ $4=0-\frac{1}{2}(9.8)t^{2}$

⇒ -4 = -4.9t² (multiply both sides by -1)

⇒ 4 = 4.9t² (divide both sides by 4.9)

⇒ 0.81633 = t² (take √ for both sides)

⇒ t = 0.9035

Then the time of the ball to land on the ground is 0.9035 seconds

The range of the ball on the ground is 30 m

The range $R=u_{x}*t$ , where $u_{x}$ is the horizontal

component of the initial velocity

R = 30 meters and t = 0.9035

⇒ $30=u_{x}(0.9035)$ (divide both sides by 0.9035)

⇒ $u_{x}=33.20$ m/s

The pitching speed of your friend is 33.20 m/s 

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2 years ago

Methods to reduce the frictional force between the an object and surface which it is in contact.

telo118 [61]

Use of lubricant
Use of ball bearers
Use of streamlined body
Use of graphite

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2 years ago

n oscillator is driven by a sinusoidal force. The frequency of the applied force A : must be less than the natural frequency of

lilavasa [31]

Answer:

B : is independent of the natural frequency of the oscillator

Explanation:

You can apply any force you like to a natural oscillator. It is independent of the natural frequency of the oscillator.

The result you get will depend on how the frequency of the applied force and the natural frequency relate to each other. It will also depend on the robustness of the oscillator with respect to the applied force.

Clearly, if the force is small enough, it will have no effect on the oscillator. If it is large enough, it will overpower any motion the oscillator may attempt. For forces in the intermediate range, there will be some mix of natural oscillation and forced behavior. One may modulate the other, for example.

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2 years ago