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frez [133]
2 years ago
8

What is the main mid west meal

Physics
1 answer:
Orlov [11]2 years ago
7 0

Answer:

Sometimes called "the breadbasket of America" the Midwest serves as a center for grain production, particularly wheat, corn and soybeans. Graham bread, steamed cornbread, and johnny cakes are traditional Midwestern foods, as are butter cakes like chocolate cake, devil's food cake, coconut cake and fruit cake

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"The White Shark" allows riders to start from rest on a tube and then slide down a 44 meter slide. It takes the rider 6.2 second
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Do any of the atom diagrams below represent atoms of the same element?
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A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo
Serggg [28]

Before the engines fail (0\le t\le3.00\,\rm s), the rocket's horizontal and vertical position in the air are

x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2

y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2

and its velocity vector has components

v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t

v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t

After t=3.00\,\rm s, its position is

x=273\,\rm m

y=362\,\rm m

and the rocket's velocity vector has horizontal and vertical components

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}

After the engine failure (t>3.00\,\rm s), the rocket is in freefall and its position is given by

x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t

y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2

and its velocity vector's components are

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}-gt

where we take g=9.80\,\frac{\rm m}{\mathrm s^2}.

a. The maximum altitude occurs at the point during which v_y=0:

159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s

At this point, the rocket has an altitude of

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve y=0 for t, then add 3 seconds to this time:

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s

So the rocket stays in the air for a total of 37.6\,\rm s.

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute x for this time t:

273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m

5 0
3 years ago
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