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pickupchik [31]
3 years ago
6

Where might Michael look to find ideas about what the teacher thinks is important information to be learned for a test? a.Vocabu

lary lists and flashcardsc.Pretests and quizzesb.His notesd.B and C only
Physics
2 answers:
masya89 [10]3 years ago
8 0
D. B and C only
My sister just took the test and im sitting next to her and D is the correct answer

Serga [27]3 years ago
3 0

Answer:

D: B and C only

Explanation:

Although vocabulary lists and flash cards may assist you in preparing for a test, if you intend to focus your studying to what the teacher thinks is important, a good idea would be to look at past papers (which can include tests that have been written before or quizzes) and your own notes.

When a teacher teaches a certain topic, they usually focus and explain the parts of that topic that they want people to understand, and if they specifically want people to understand one certain section, there is a high chance that they will put that topic in the test.

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Climates on Earth get _____ as you move from the equator to the poles.
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Explanation:

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3 years ago
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A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at
Vesna [10]

Answer:

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0  So B=0

-16 B-32 A+16B=6  So A=-3/16

y=-3/16 cos 4t

Now to find the CF  of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

m^2+8m+16=0

So CF =(C_1+tC_2)e^{-2t}

So the general equation

Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t

Given that t=0 Y=0 So

C_1=\dfrac{3}{16}

t=0 Y'=0 So

C_2 =\dfrac{3}{8}

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

The above equation is the general equation for motion.

3 0
3 years ago
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