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4 years ago

Compare erosion and deposition.

Physics

2 answers:

pishuonlain [190]4 years ago

6 0

Erosion makes rocks wear away and deposition deposits sediment.

hoa [83]4 years ago

6 0

Erosion is the move ment of sedement deposison is the depoentment of sediment

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A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk

Murrr4er [49]

Answer:

$\omega_2=5.1rad/s$

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is $L=I\omega$ . If we call 1 the situation when the student is at the rim and 2 the situation when the student is at $r_2=1.39m$ from the center, then we have:

$L_1=L_2$

Or:

$I_1\omega_1=I_2\omega_2$

And we want to calculate:

$\omega_2=\frac{I_1\omega_1}{I_2}$

The total moment of inertia will be the sum of the moment of intertia of the disk of mass $m_D=119.1 kg$ and radius $r_D=3.23m$ , which is $I_D=\frac{m_Dr_D^2}{2}$ , and the moment of intertia of the student of mass $m_S=54.3kg$ at position $r$ (which will be $r_1=r=3.23m$ or $r_2=1.39m$ ) will be $I_{S}=m_Sr_S^2$ , so we will have:

$\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}$

or:

$\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}$

which for our values is:

$\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s$

6 0

3 years ago

Beginning 156 miles directly east of the city of Uniontown, a truck travels due south. If the truck is travelling at a speed of

timurjin [86]

Answer:

14.3

Explanation:

The distance s as a function of time can be written as:

$s(t) = \sqrt{156^{2} + (31t)^2}$

The rate of change is the derivative of d with respect to time:

$\frac{ds}{dt} =\frac{961t}{\sqrt{156^{2}+(31t)^{2}}}$

The time t when the track has been traveling for 81 miles is given by:

$81 = 31t\\ t = \frac{81}{31}$

Using t in the previous equation gives:

$\frac{ds}{dt}(\frac{81}{31} ) =\frac{2511}{\sqrt{156^{2}+81^{2}}}=14.3$

5 0

3 years ago

What is Obama's last name? PLZ PLZ PLZ THIS IS 500% OF MY GRADE PLS HELP PLS

kotegsom [21]

Answer:

Barack Hussein Obama II

Explanation:

I think

4 0

3 years ago

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A 2.0 kg body rests on a frictionless wedge that has an inclination of θ = 35° and an acceleration to the right such that the ma

quester [9]

OK, the wedge is accelerating (a) at Theta = 180 degrees (to the right) and the wedge is inclined theta = 75 degrees. For the m = 2 kg block to remain at rest all we need is a net force f = W cos(theta) - F sin(theta) = 0; where F = ma and W = mg the weight of the block. That is, the weight component along the incline is offset by the acceleration component along the surface; so the block does not slide.

Solving we have W cos(theta) = mg cos(theta) = ma sin(theta) = F sin(theta); such that a = g cos(theta)/sin(theta) = g cot(theta). Assuming g ~ 9.81 m/sec^2, you can now plug and chug to find the answer.

<span>The physics is this...when the net force on a body is f = 0, that body will not accelerate and start to move if it is already still. So when the block's weight component along the surface of the wedge is offset by the equal but opposite force along the surface of the accelerating wedge, the still block will not move.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>

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4 years ago

PLEASE HELP!!!!

Gala2k [10]

<span>Energy of a shear wave relates to the direction of a particle motion is downwards.</span>

3 0

3 years ago

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