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3 years ago

An inclined plane can help _____.

Physics

1 answer:

mylen [45]3 years ago

5 0

Reduce the effort needed to move a load vertically

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A potter's wheel rotating at 215 rev/min is switched off and rotates through 75.0 revolutions prior to coming to rest. What is t

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Explanation:

Given :

Initial angular velocity =w_i= 215 rev/min

Final angular velocity =w_f= 0

Number of revolutions=n =75.0 rev

Using relation

$w_f^2-w_i^2=2\alpha n$

$0-215^2=2\times 75\alpha$

$-308.1666 rev/min^2=\alpha$

Converting to rad/s^2

$\alpha =-1.45 rad/s^2$

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3 years ago

which of these waves requires a medium to travel through light wave electromagnetic waves sound wave microwave

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Sound waves require a medium to travel through. :D

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4 years ago

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3 years ago

A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b

guapka [62]

<h2>Answer: (a)t=0.553s, (b)x=110.656m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=200m/s$ is the bullet's initial speed

$\theta=0$ because we are told the bullet is shot horizontally

$t$ is the time since the bullet is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the bullet

$y=0$ is the final height of the bullet (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:

$0=1.5m+200m/s.sin(0) t-\frac{9.8m/s^{2}.t^{2}}{2}$ (3)

$0=1.5m-\frac{9.8m/s^{2}.t^{2}}{2}$ (4)

Finding $t$ :

$t=\sqrt{\frac{1.5m(2)}{9.8m/s^{2}}}$ (5)

Then we have the time elapsed before the bullet hits the ground:

$t=0.553s$ (6)

For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:

$x=V_{o}cos\theta t$ (1)

Substituting the knonw values and the value of $t$ found in (6):

$x=200m/s.cos(0)(0.553s)$ (7)