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timofeeve [1]
3 years ago
8

Suppose that an owner of the same dog breed has also taken some measurements. They notice that the surface area of the dog has i

ncreased by a factor of 3 over a period of four years. Determine the change in the dog's relative surface area in this time period.
Physics
1 answer:
Mnenie [13.5K]3 years ago
7 0

Answer:

Surface area of the dog is changes from A to 3A

Explanation:

It is given that surface area of dog is increased by factor 3 in a period of 4 year

We have to find the change in dog's relative surface area in the given time period

Let initially the surface area is A

As the surface area is increased by a factor of 3

So surface area after 4 year = 3×A = 3A

So surface area of the dog is changes from A to 3A

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A 5.7 kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0s, the bl
Genrish500 [490]

Answer:

Explanation:

Given

mass of block m=5.7\ kg

at t=0 s

displacement is x=-0.7\ m

velocity v=-0.8\ m/s

acceleration a=2.7\ m/s^2

suppose x=A\cos (\omega t+\phi )   is the general equation of SHM

where A=amplitude

\omega=natural frequency of oscillation

therefore velocity and acceleration is given by

v=-A\omega \sin (\omega t+\phi )

a=A\omega ^2\cos (\omega t+\phi )

for t=0

-0.7=A\cos (\phi )---1

v=-0.8=-A\omega \sin(\phi)---2

a=2.7=-A\omega ^2\cos(\phi )----3

divide 1 and 3 we get

\omega ^2=\frac{27}{7}

\omega =\sqrt{\frac{27}{7}}

Now square and 1 and 2 we get

(0.7)^2+(\frac{0.8}{\omega })^2=A^2

A^2=0.49+0.166

A=0.81\ m

     

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Explanation:

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