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telo118 [61]
2 years ago
14

The hill is covered in gravel so that the truck's wheels will slide up the hill instead of rolling up the hill. The coefficient

of kinetic friction between the tires and the gravel is k. This design has a spring at the top of the ramp that will help to stop the trucks. This spring is located at height h. The spring will compress until the truck stops, and then a latch will keep the spring from decompressing (stretching back out). The spring can compress a maximum distance x because of the latching mechanism. Your job is to determine how strong the spring must be. In other words, you need to find the spring constant so that a truck of mass mt, moving at an initial speed of v0, will be stopped. For this problem, it is easiest to define the system such that it contains everything: Earth, hill, truck, gravel, spring, etc. In all of the following questions, the initial configuration is the truck moving with a speed of v0 on the level ground, and the final configuration is the truck stopped on the hill with the spring compressed by an amount x. The truck is still in contact with the spring. Solve all of the questions algebraically first. Then use the following values to get a number for the desired answer.

Physics
1 answer:
mamaluj [8]2 years ago
8 0

Answer:

Explanation:

check attachment  for the solution.

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Neurons that deliver sensory information from sensory receptors to the spinal cord are called __________.
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Answer:

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Explanation:

First Order Neurons

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An electron and a proton have the same kinetic energy upon entering a region of constant magnetic field and their velocity vecto
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Answer: rp/re= me/mp= 544 * 10^-6.

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Then the dynamic equation for the circular movement is given by:

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we write this for each particle then we have the following:

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Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get
olga_2 [115]

Answer:

Part a)

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Part b)

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Part c)

\alpha = 0.69 rad/s^2

Part d)

\alpha = 0.48 rad/s^2

Part e)

t = 9.14 s

Explanation:

Part a)

Angular speed is given as

\omega = 2\pi f

\omega = 2\pi(\frac{78}{60})

\omega = 8.17 rad/s

Part b)

Since turn table is accelerating uniformly

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\theta = \frac{\omega_f + \omega_i}{2} t

\theta = \frac{8.17 + 0}{2}(11.9)

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Part c)

angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{t}

\alpha = \frac{8.17 - 0}{11.9}

\alpha = 0.69 rad/s^2

Part d)

When its angular speed changes to 120 rpm

then we will have

\omega_2 = 2\pi (\frac{120}{60})

\omega_2 = 12.56 rad/s

number of turns revolved is 15 times

so we have

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

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Part e)

now for uniform acceleration we have

\omega_f - \omega_i = \alpha t

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t = 9.14 s

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