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3 years ago

Math short answer question! Please help!

Mathematics

1 answer:

IRISSAK [1]3 years ago

7 0

Part 1:

The circumference of the smaller circle is...
C = 2*pi*r
C = 2*pi*2
C = 4*pi

The circumference of the larger circle is...
C = 2*pi*r
C = 2*pi*8
C = 16*pi

The ratio of the two circumferences is
16*pi:4*pi = 4:1
which means that for every single full rotation of the larger gear, the smaller gear will complete four rotations

If we flip things around, we will have the ratio 4:1 turn into 1:0.25 (divide both parts by 4). Now this says that for every single rotation of the smaller gear, the larger gear rotates 1/4 or 25% of the larger circle circumference.

1/4 of 360 degrees is 90 degrees

So the larger gear will rotate 90 degrees when the small gear completes a full revolution.

Answer: 90 degrees

=====================================================

Part 2:

This was pretty much answered in part 1 above. The answer here is 4. The circumference of the larger gear is 4 times larger (since the radius is 4 times larger), so the gear ratio here is 4:1.

Answer: 4

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Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

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Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

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