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2 years ago

Calculate the frequency of the wave shown below.

Physics

1 answer:

jenyasd209 [6]2 years ago

5 0

$\color{skyblue}{ \underline{ \frak { \: option \: ( \: c \: ) = 2 \: hertz ✓}}}$

$\: \:$

Given :

Wavelength ( λ ) = 2 m

$\: \:$

Speed = 4 m/s

$\: \:$

We, have to find frequency :

$\:$

$\large \tt \: Frequency = \frac{Speed}{Wavelength ( \: λ \: )}$

$\: \:$

$\large \tt \: Frequency = \frac{4}{2}$

$\: \:$

$\large \tt \: Frequency = \cancel \frac{4}{2}$

$\: \:$

$\pink{ \boxed{\large \tt \: Frequency =2 \: Hertz ✓}}$

$\: \:$

Hope Helps!

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3 years ago

A rock climber wears a 8.1 kg backpack while scaling a cliff. After 28.2 min, the climber is 9.4 m above the starting point. How

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Answer:

Explanation:

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weight of climber, W = 656 N

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time, t = 28.2 min = 28.2 x 60 = 1692 second

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4 years ago

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The chair does not move along the vertical direction: this means that the net force on the vertical direction is zero.

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- The component of the force F perpendicular to the floor, acting downward: $F_y = F sin 43.0^{\circ}=(44.0 N)(sin 43.0^{\circ})=30.0 N$

- The normal force N, acting upwards.

Since the net force must be zero, we have:

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From which we can find the value of N:

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4 years ago

Read 2 more answers

que 2. Why do we keep frequency constant instead of keeping vibrating length constam second law of vibrating string?

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Answer:

The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant

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$f = \dfrac{1}{2\cdot l} \cdot \sqrt{\dfrac{T}{m} }$

The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously

Therefore, the law is verified indirectly, by rearranging the above equation as follows;

$m = \dfrac{1}{ l^2} \cdot \dfrac{T}{4\cdot f^2} }$

From which it can be shown that the following relation holds with the limits of error in the experiment

m₁·l₁² = m₂·l₂² = m₃·l₃² = m₄·l₄² = m₅·l₅²

Explanation:

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