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Andre45 [30]
3 years ago
6

A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.

Physics
1 answer:
Crank3 years ago
8 0

Answer: 1000 Hz

Explanation:

You can calculate frequency by dividing velocity by wavelength

Frequency = velocity/wavelength

Find velocity first.

900 m/3 s = 300 m/s

Plug values in to find frequency.

F = (300 m/s)/0.3 m

F = 1000 Hz

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A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with
NNADVOKAT [17]

Answer: 1.59 sec

Explanation:

The kinetic energy contained in the rotation of the cylinder is 1/2 m v^2

The kinetic energy of translation is also 1/2 m v^2

so the total energy is m v^2

The force applied is mg sin (theta)

= m x 9.8 x 1/2

= 4.9 m

Now equate

F x d = m v^2

4.9 m x 3.1 = m v^2

v^2 = 4.9 x 3.1

v = sqrt(4.9 x 3.1) = 3.9 m/s

Acceleration.

V^2 = 2 a s

4.9 x 3.1 = 2 x a x 3.1

4.9 = 2 a

a = 2.45 m/s^2

Time

T = v/a

= 3.9/2.45

= 1.59 sec

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4 years ago
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How does temperature,barometric pressure,humidity,wind speed and direction, and precipitation determine the weather in a particu
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4 years ago
The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

3 0
4 years ago
Read 2 more answers
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