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mafiozo [28]
3 years ago
15

When the source of a sound is moving it's speed increases

Physics
2 answers:
Novay_Z [31]3 years ago
7 0
That statement is false, and pretty meaningless ... we don't know
whether "its speed" refers to the source or the sound.

When the source of a sound is moving relative to the listener, then
the listener may hear a higher- or lower-pitch sound than what the
source is actually emitting.
Burka [1]3 years ago
5 0

Answer:

false

Explanation:

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Everything would be more advanced and logical while the past people really just tried to survive.
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Did the heliocentric model explain the drawbacks of the geocentric model?
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Using a rope that will snap if the tension in it exceeds 387 N, you need to lower a bundle of old roofing material weighing 449
Anni [7]

Answer:

(a) a\approx1.4 m.s^{-2}

(b) v\approx 4.133 m.s^{-1}

Explanation:

Given:

  • Limiting tension of snapping of the rope, T= 387 N
  • Weight of the object to be lifted, w=449 N
  • ∴mass, \Rightarrow m= 45.8163 kg
  • height of letting down the weight, h = 6.1 m

(a)

Now,

The force to be compensated for  being on the verge of snapping:

(T-w) = 62 N

Therefore, we need to produce and acceleration equivalent to the above force.

∵F=m.a

62=45.8163\times a

a= \frac{62}{45.8163}

a\approx 1.4 m.s^{-2}

(b)

From the equation of motion ,we have:

v^{2} =u^{2} +2a.s....................(2)

where:

u= initial velocity= 0 (here, starting from rest)

v= final velocity = ?

a= 1.4 m.s^{-2}

s= displacement =h =6.1 m

Now, putting the values in eq. (2)

v^2= 0^2 + 2\times 1.4\times 6.1

v\approx 4.133 m.s^{-1}        is the velocity with which the body will hit the ground in the given conditions.

7 0
4 years ago
A still ball of mass 0.514kg is fastened to a cord 68.7cm long and is released when the cord is horizontal. At the bottom of its
Alex787 [66]

Answer:

1.21 m/s

Explanation:

From the law of conservation of energy,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

where U₁ = initial potential energy of system =initial potential energy of still ball = mgh where m = mass of still ball = 0.514 kg, g = acceleration due to gravity = 9.8 m/s² and h = height = length of cord = 68.7 cm = 0.687 m.

K₁ = initial kinetic energy of system = 0

E₁ = initial internal energy of system = unknown and

U₂ = final potential energy of system = 0

K₁ = final kinetic energy of system = final kinetic energy of ball + steel block = 1/2(m + M)v² where m = mass of still ball, M = mass of steel block = 2.63 kg and v = speed of still ball + steel block

E₁ = final internal energy of system = unknown

So,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂

mgh = 1/2(m + M)v² + (E₂ - E₁)

Given that (E₂ - E₁) = change in internal energy = ΔE = 1/2ΔK where ΔK = change in kinetic energy. So, ΔE = 1/2ΔK = 1/2(K₂ - K₁) = K₂/2 = 1/2(m + M)v²/2 = (m + M)v²/4

Thus, mgh = 1/2(m + M)v² + (E₂ - E₁)

mgh = 1/2(m + M)v² + (m + M)v²/4

mgh = 3(m + M)v²/4

So, making v subject of the formula, we have

v² = 4mgh/3(m + M)

taking square root of both sides, we have

v = √[4mgh/3(m + M)]

Substituting the values of the variables into the equation, we have

v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/{3(0.514 kg + 2.63 kg)}]

v = √[13.8422/{3(3.144 kg)}]

v = √[13.8422 kgm/s²/{9.432 kg)}]

v = √(1.4676 m²/s²)

v = 1.21 m/s

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Pavlova-9 [17]

Answer:

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