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weeeeeb [17]
3 years ago
7

While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find th

e motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 16 m , and you use your watch to find that each loop around takes 29 s. What is the speed?
Physics
1 answer:
Zepler [3.9K]3 years ago
5 0

Answer:

v=3.47m/s

Explanation:

The speed is by definition the distance traveled divided over the time it takes to travel that distance. In this case, this distance is the circumference of the wheel, so we have:

v=\frac{C}{t}=\frac{2\pi r}{t}

where we have written the circumference in terms of its radius.

For our values we then obtain the value:

v=\frac{2\pi r}{t}=\frac{2\pi (16m)}{(29s)}=3.47m/s

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3 years ago
6. A light ray strikes a reflective plane surface at an angle of 560 with the surface.
Zolol [24]

Answer:

deez nouts

Explanation:

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3 years ago
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pogonyaev

Answer:

3a, 2b,4c,1d

Explanation:

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7 0
3 years ago
In odd nuclei, what determines the final spin of the nucleus?
Katen [24]

Answer:

In odd nuclei, the left out proton or neutron will contribute to the spin of the nucleus.

Explanation:

The meaning of odd nuclei is atomic mass is odd.

A=odd number.

A=Z+n

Here, Z is proton either it will odd or n will odd which is neutron.

Now according to the shell model the left out proton or neutron will contribute to the spin and parity.

For example,

Take the case of isotope of nitrogen-15.

Here Z is 7, and n is 8 will not contribute in spin.

Now, for Z=7.

1S^{2} _{\frac{1}{2} }, 1P^{4} _{\frac{3}{2} }, 1P^{1} _{\frac{1}{2} }

Here,

j=\frac{1}{2}

and, L=1.

Fort parity,

(-1)^{L}

Put the value of L.

Parity will be -1.

Now, spin will be

S=(\frac{1}{2} )^{-1}.

7 0
3 years ago
81. A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a
Irina18 [472]

Answer:

Explanation:

Given

mass of squirrel m=560 gm

Surface area of squirrel A=930 cm^2

and the area which face A_f=\frac{A}{2}=465 cm^2

height of tree h=5 m

Coefficient of drag C=1

drag Force F_d=\frac{1}{2}C\cdot \rho \cdot A\cdot v^2

Terminal velocity is given

F_d=mg

\frac{1}{2}C\cdot \rho \cdot A\cdot v^2=mg

v=\sqrt{\frac{2\times m\times g}{\rho \times C\times A_f}}

v=\sqrt{\frac{2\times 0.560\times 9.8}{1.2\times 1\times 465\times 10^{-4}}}

v=13.9 m/s

(b)Mass of person m=56 kg

v^2-u^2=2gh

u=0

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 5}

v=9.89 m/s

7 0
3 years ago
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