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ElenaW [278]
4 years ago
8

A wave has a speed of 30 m/s, a frequency of 6 Hz, and a wavelength of 5 m. If the wavelength remains constant, and the frequenc

y is doubled, what is the new speed of the wave?
Physics
1 answer:
mixer [17]4 years ago
8 0

Answer: 60m/s

Explanation:

The wavespeed is the distance covered by the wave in one second. It is measured in metre per second, and represented by the symbol V

Wavespeed (V) = Frequency F x wavelength λ

i.e V = F λ

In the first case:

Wavespeed = 30 m/s

Frequency of sound = 6Hz

Wavelength = 5m

In the second case:

Wavespeed = ?

Frequency of sound = (2x 6Hz = 12Hz)

Wavelength = 5m (remains constant)

Apply V = F λ

Wavespeed = 12 Hz x 5m

Wavespeed = 60m/s

Therefore, when frequency is doubled, the speed is also doubled. Thus, the new speed of the wave is 60m/s

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Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the
yaroslaw [1]

Answer:

 K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

          p₀ = 0

final moment

          p_f = m_a v_a + m_b v_b

          p₀ = p_f

          0 = m_a v_a + m_b v_b

tells us that

          m_a = 8 m_b

         

           0 = 8 m_b v_a + m_b v_b

           v_b = - 8 v_a                    (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

           Em₀ = K₀ = 73 J

final point. After separating the body

          Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

           K₀ = K_f

           73 = ½ m_a (v_a² + v_b² / 8)

           

we substitute equation 1

           73 = ½ m_a (v_a² + 8² v_a² / 8)

           73 = ½ m_a (9 v_a²)

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3 0
3 years ago
At a hot air balloon race, a person on the ground shots a ball of confetti from a cannon to start the race. One of the balloons
Sati [7]

Answer:

a) The balloon and the ball will meet after 1.43 s.

b) The ball will reach the balloon at 15.7 m above the ground.

Explanation:

The height of the confetti ball is given by the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

The height of the ball is given by this equation:

y = y0 + v · t

Where v is the constant velocity.

When the ball and the ballon meet, both heights are equal. Let´s consider the ground as the origin of the frame of reference so that y0 = 0:

y balloon = y ball

y0 + v · t = y0 + v0 · t + 1/2 · g · t²                  (y0 = 0)

11 m/s · t = 18 m/s · t -1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m/s · t

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0 = t( -4.9 m/s² · t  + 7 m/s)

t = 0 and

0 = -4.9 m/s² · t  + 7 m/s

-7 m/s / - 4.9 m/s² = t

t = 1.43 s

They will meet after 1.43 s

b) Now let´s calculate the height of the balloon after 1.43 s

y = v · t

y = 11 m/s · 1.43 s = 15.7 m

The ball will reach the balloon at 15.7 m above the ground.

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Answer:

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