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arlik [135]
2 years ago
8

If one end of a heavy rope is attached to one end of a lightweight rope, a wave can move from the heavy rope into the lighter on

e. (ii) What happens to the frequency? Choose from the same possibilities. (a) It increases.(b) It decreases.(c) It is constant.(d) It changes unpredictably.
Physics
1 answer:
Kamila [148]2 years ago
3 0

We can actually deduce here that if one end of a heavy rope is attached to one end of a lightweight rope, a wave can move from the heavy rope into the lighter one. What happens to the frequency that it increases.

Thus, option A. It increases is the answer.

<h3>What is frequency?</h3>

Frequency is actually known to be the amount or number of waves that passes a fixed point per unit time. In physics, frequency is also known as the number of cycles or vibrations per unit time.

When the wave move from the heavy rope into the lighter one, the wave speed will definitely increase. This is because when a wave is created from the end of the heavy rope, the frequency will start small and then increases as it gets to the end of the lighter rope.

Learn more about frequency on brainly.com/question/254161

#SPJ4

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an object is moving with initial velocity of 5 m/s. After 10 seconds final velocity is 10 m/s. Calculate its acceleration.​
Brut [27]

Answer:

0.5m/s2

Explanation:

acceleration= change in velocity/time taken

= v - u/ t

= 10-5/10

=5/10

= 0.5m/s2

6 0
3 years ago
A space shuttle sits on the launch pad for 2.0 minutes, and then goes from rest to 4600 m/s in 8.0 minutes. Treat its motion as
SpyIntel [72]

Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

    Acceleration (a) is defined as the time rate of change of velocity

                       a = \frac{v_{2} - v_{1} } {t}  

Given data

 Final velocity = v₂ = 0 m/s

 Initial velocity = v ₁ = 0 m/s

  As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

                 a = 0 ms⁻²

b.)

     Given data

As the space shuttle start from rest, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 8 min = 480 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {480}

                     a = 9.58 ms⁻²

c.)

    Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 10 min = 600 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {600}

                     a = 7.67 ms⁻²

8 0
3 years ago
Peak to peak value of voltage is
hram777 [196]

Answer: Peak-to-peak voltage is the distance from the lowest negative amplitude, or trough, to the highest positive amplitude, or crest, of the AC voltage waveform. In other words, peak-to-peak voltage is equal to the full height of the waveform. Peak-to-peak voltage can be found using peak voltage or RMS voltage.

Explanation:hope tht gave u a clue have a wonderful Christmas Eve time with your family!❄️

3 0
2 years ago
Read 2 more answers
Light travels at the same speed at all times. True False
topjm [15]

Answer:

true hope this help :}

Explanation:

6 0
3 years ago
Read 2 more answers
A pendulum on earth swings with angular frequency ω. On an unknown planet, it swings with angular frequency ω/ 4. The accelerati
Afina-wow [57]

Answer:

g / 16

Explanation:

T = 2π \sqrt{\frac{l}{g} }

angular frequency ω = 2π /T

= \sqrt{\frac{g}{l} }

ω₁ /ω₂ = \sqrt{\frac{g_1}{g_2} }

Putting the values

ω₁ = ω ,     ω₂ = ω / 4

ω₁ /ω₂ = 4

4 =  \sqrt{\frac{g}{g_2} }

g₂ = g / 16

option d is correct.

6 0
3 years ago
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