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timurjin [86]
3 years ago
14

Bob and Lily are riding on a typical carousel. Bob rides on a horse near the outer edge of the circular platform, and Lily rides

on a horse near the center of the circular platform. When the carousel is rotating at a constant angular speed, Bob's angular speed is
Physics
1 answer:
alukav5142 [94]3 years ago
6 0

Answer:

Bob's angular speed is the same as that of lily

Explanation:

Because for a carousel the angular speed remains the same since velocity at center and edge are the same

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The electric field strength in the space between two closely spaced parallel disks is 1.0 10^5 N/C. This field is the result of
alex41 [277]

To solve this problem it is necessary to apply the concepts related to the capacitance in the disks, the difference of the potential and the load in the disc.

The capacitance can be expressed in terms of the Area, the permeability constant and the diameter:

C = \frac{\epsilon_0 A}{d}

Where,

\epsilon_0 = Permeability constant

A = Cross-sectional Area

d = Diameter

Potential difference between the two disks,

V = Ed

Where,

E = Electric field

d = diameter

Q = Charge on the disk equal to \rightarrow Q=ne=(3.9*10^9)(1.6*10^{-19})= 6.24*10^{-10}C

Through the value found and the expression given for capacitance and potential, we can define the electric charge as

Q = CV

Q = \frac{\epsilon A}{d}(Ed)

Q = \epsilon_0 AE

Q = \epsilon_0 \pi(\frac{d}{2})^2E

Q = \frac{\epsilon \pi d^2E}{4}

Re-arranging the equation to find the diameter of the disks, the equation will be:

d = \sqrt{\frac{4D}{\epsilon_0 \pi E}}

Replacing,

d = \sqrt{\frac{4(6.24*10^{-10})}{(8.85*10^{-12})\pi(1*10^{5})}}

d = 0.0299m

Therefore the diameter of the disks is 0.03m

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3 years ago
What would you do to improve the precision of an experiment?
GenaCL600 [577]

Explanation:

Precision represents that how close the different measurements of the sample one take are to one another.

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A). Both the energy and the wave travel in the same direction.

If they didn't, they'd wind up in different cities almost instantly.

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A student connects an object with mass m to a rope with a length r and then rotates the rope around her head parallel to the gro
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The object takes 0.5 seconds to complete one rotation, so its rotational speed is 1/0.5 rot/s = 2 rot/s.

Convert this to linear speed; for each rotation, the object travels a distance equal to the circumference of its path, or 2<em>π</em> (1.2 m) = 2.4<em>π</em> m ≈ 7.5 m, so that

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Then the tension in the rope is

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