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marta [7]
3 years ago
12

This problem uses the same concepts as Multiple-Concept, except that kinetic, rather than static, friction is involved. A crate

is sliding down a ramp that is inclined at an angle 30.6 ° above the horizontal.
The coefficient of kinetic friction between the crate and the ramp is 0.360. Find the acceleration of the moving crate.
Physics
1 answer:
Bingel [31]3 years ago
3 0

Answer:

1.96 m/s^2

Explanation:

angle of inclination, θ = 30.6°

coefficient of friction, μ = 0.36

Acceleration of the crate

a = g Sin θ - μ g Cos θ

a = 9.8 ( Sin 30.6 - 0.36 x Cos 30.6)

a = 9.8 ( 0.509 - 0.309)

a = 1.96 m/s^2

Thus, the acceleration f the crate is 1.96 m/s^2.

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ANSWER: Destructive Interference

-This is the exact definition of the question you have provided, look this term up if you do not believe lol.

Hope this Helps!
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Question 4 What would be the UCS for Tab the dog? O The shock O The flowers and fence O Fear O The boundary Question 5​
Harrizon [31]

The unconditioned stimulus (UCS) for Tab the dog is: B. The flowers and fence.

<h3>The types of stimuli.</h3>

In Science, the two (2) stimuli that are repeatedly paired in classical conditioning include:

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  • Unconditioned stimulus

<h3>What is an unconditioned stimulus?</h3>

An unconditioned stimulus can be defined as a stimulus that is capable of naturally triggering a response, before or without any conditioning while it elicit a specific response without learning.

  • This ultimately implies that, an unconditioned stimulus leads to an automatic response in a living organism such as a dog.

In this scenario, the unconditioned stimulus (UCS) for Tab the dog is simply the flowers and fence.

Read more on unconditioned stimulus here: brainly.com/question/24868138

3 0
2 years ago
What type of bulb is unaffected ​
emmasim [6.3K]
<h2>A N S W E R : –</h2>

  • c) Parallel

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3 0
2 years ago
If the load on Pulley E is 20N, how much effort is needed to life it?
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7 0
3 years ago
A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity
nadezda [96]

Answer:

u(t)=\frac{1}{5} sin\ (25t)

Explanation:

Given:

  • mass of the body stretching the spring, m=150\ g
  • extension in spring, \Delta x=1.568\ cm
  • velocity of oscillation, u'(0)=20\ cm.s^{-1}
  • initial displacement position of equilibrium, u(0)=0

<u>According to given:</u>

m.g=k.\Delta x

150\times 980=k\times 1.568

k=93750\ dyne.cm^{-1}

<u>we know frequency:</u>

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{93750}{150} }

\omega=25

Now, for position of mass in oscillation:

u= A.sin\ (\omega.t)+B.cos\ (\omega.t)

u= A.sin\ (25.t)+B.cos\ (25.t)

at t=0;\ u(0)=0\ \Rightarrow A=0

∴u(t)=B.sin\ (25.t)

∵ at t=0;\ u'(0)=20\ cm.s^{-1}\ \Rightarrow B=\frac{1}{5}

u(t)=\frac{1}{5} sin\ (25t)

7 0
3 years ago
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