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rodikova [14]
3 years ago
9

A busy chipmunk runs back and forth along a straight line of acorns that has been set out between its burrow and a nearby tree.

At some instant, it moves with a velocity of −1.09 m/s−1.09 m/s . Then, 2.99 s2.99 s later, it moves with a velocity of 1.75 m/s1.75 m/s . What is the chipmunk's average acceleration during the 2.99 s2.99 s time interval
Physics
1 answer:
Nadusha1986 [10]3 years ago
5 0

Answer:

0.59 seconds

Explanation:

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A rock has a mass of 3.1 kg. What is its weight on earth
Masteriza [31]

Answer:

W = 30.38 N

Explanation:

Given that,

Mass of a rock, m = 3.1 kg

We need to find the weight of the rock on the surface of Earth. Weight of an object is given by :

W = mg

g is the acceleration due to gravity, g = 9.8 m/s²

W = 3.1 kg × 9.8 m/s²

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So, the weight of the rock on the Earth is 30.38 N.

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Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat
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Answer:

a) a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

b) k = \frac{9.8}{9.74}=1.006

Explanation:

Part a

For this case we can begin finding the period like this:

T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s

Then we know that the centripetal acceleration is given by:

a= \frac{v^2}{r}

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v=\frac{2\pi r}{T}

If we replace this into the acceleration we got:

a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}

And we can replace the values and we got:

a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

Part b

For this case we want to find a value of k such that:

a= k 9.8

Where a = 9.74, so then we can solve for k like this:

k = \frac{9.8}{9.74}=1.006

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Answer:

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just search it up

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