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iren [92.7K]
3 years ago
6

A long ramp made of cast iron is sloped at a constant angle θ = 52.0∘ above the horizontal. Small blocks, each with mass 0.42 kg

but made of different materials, are released from rest at a vertical height h above the bottom of the ramp. In each case the coefficient of static friction is small enough that the blocks start to slide down the ramp as soon as they are released. You are asked to find h so that each block will have a speed of 4.00 m/s when it reaches the bottom of the ramp. You are given these coefficients of sliding (kinetic) friction for different pairs of materials.
Material 1 Material 2 Coefficient of Sliding Friction
Cast iron Cast iron 0.15
Cast iron Copper 0.29
Cast iron Lead 0.43
Cast iron Zinc 0.85
Physics
1 answer:
dezoksy [38]3 years ago
4 0

Answer:

For cast iron we have

h = 0.92 m

For copper

h = 1.05 m

For Lead

h = 1.23 m

For Zinc

h = 2.43 m

Explanation:

As we know that final speed of the block is calculated by work energy theorem

W_f + W_g = \frac{1}{2}mv^2

now we have

-\mu_k mg cos\theta(\frac{h}{sin\theta}) + mgh = \frac{1}{2}mv^2

now we have

v^2 = 2gh - 2\mu_k g h cot\theta

v = \sqrt{2gh(1 - \mu_k cot\theta)}

For cast iron we have

4 = \sqrt{2(9.81)(h)(1 - 0.15cot52)}

h = 0.92 m

For copper

4 = \sqrt{2(9.81)(h)(1 - 0.29cot52)}

h = 1.05 m

For Lead

4 = \sqrt{2(9.81)(h)(1 - 0.43cot52)}

h = 1.23 m

For Zinc

4 = \sqrt{2(9.81)(h)(1 - 0.85cot52)}

h = 2.43 m

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A student releases a marble from the top of a ramp. The marble increases speed while on the ramp then continues across the floor
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3 years ago
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A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a
jeka94

Answer:

a)  t_l - t_r = 12.54 us

b)  (t_l - t_r) / T = 0.0157  

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

a. What is the difference in the arrival time of the sound at the left ear and the right ear?

b. What is the ratio of this time difference to the period of the sound wave?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

                                      d_l = sqrt ( 2.6^2 + 0.15^2)

                                      d_l = sqrt ( 6.7825 )

                                      d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

                                      t_l - t_r = (1 / v) * ( d_l - d_r)

Where, v = 343 m/s speed of sound in air:

                                      t_l - t_r = (1 / 343) * ( 2.6043 - 2.6)  

                                      t_l - t_r = ( 0.0043 / 343 )

                                      t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

                                      T = 1 / f

                                      T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

                                      (t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

                                      (t_l - t_r) / T = 0.0157  

3 0
3 years ago
A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli
fredd [130]

Answer:

Part a)

A = 1.78 m

Part b)

f = 2 rev/s

Explanation:

Part A)

As we know that time period of the motion is given as

T = 2.68 s

so we have

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{2.68}

\omega = 2.34 rad/s

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

mg = m\omega^2 A

so we have

A = \frac{g}{\omega^2}

A = \frac{9.81}{2.34^2}

A = 1.78 m

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

mg = m\omega^2 A

g = \omega^2 (0.0623)

\omega = 12.5 rad/s

so now we have

2\pi f = 12.5

f = 2 rev/s

3 0
3 years ago
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