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Reika [66]
2 years ago
10

A student tries to produce notes of higher frequency by blowing a trumpet harder. he will succeed?​

Physics
2 answers:
sdas [7]2 years ago
7 0

Answer:

Please mark as Brainliest!!

Explanation:

He will not succeed as it blows the trumpet harder that only increase the amplitude of the sound wave that means the intensity of the sound will increase but the frequency remains the same.

melamori03 [73]2 years ago
3 0

Answer:

he wont

Explanation:

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a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
lilavasa [31]

Answer:

D/H =15

Explanation:

  • We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
  • At this point, we can find the value of H, applying the following kinematic equation:

       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
  • When we replace these values in (1), we find that  v₁ = -v₀.
  • Replacing in (3), we have:

        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

3 0
3 years ago
What is the momentum of an object with 5.22 kg and 1.7 m/s
BlackZzzverrR [31]

Answer:

8.874

Explanation:

You need to times 5.22 kg and 1.7 m/s to get 8.874.

8 0
3 years ago
A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two p
Anettt [7]

Answer:

A   u = 0.36c      B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

     u ’= (u-v) / (1- uv / c²)

Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

      u’ (1- u v / c²) = u -v

      u + u ’uv / c² = v - u’

      u (1 + u ’v / c²) = v - u’

      u = (v-u ’) / (1+ u’ v / c²)

Let's calculate

      u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)

      u = 0.51c / (1 + 0.4042)

      u = 0.36c

We repeat the calculation for the other piece

In this case u ’= - 0.35c

We calculate

       u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)

       u = 1.29c / (1- 0.329)

       u = 0.961c

6 0
3 years ago
Give three examples of objects in equilibrium found in classroom?​
Charra [1.4K]

Answer:

Book. Bottle. table are some examples of objects in equilibrium found in classroom

7 0
2 years ago
What are higher-than-usual tides that form when the suns and moons gravitational pulls work together
Soloha48 [4]
These are called neap tides!

Hope this helps :)
8 0
3 years ago
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