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Physics

1 answer:

user100 [1]3 years ago

4 0

Answer:

D is not the a vector quantities

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A record player has a velocity of 33.33 RPM. How fast is the record spinning in m/s at a distance of 0.085 m from the center?

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Answer:

A record player has a velocity of 33.33 RPM. How fast is the record spinning in m/s at a distance of 0.085 m from the center? [0.297 m/s] 6. A merry-go-round a.k.a “the spinny thing” is rotating at 15 RPM, and has a radius of 1.75 m A.

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If a rod attached to the approaching charge if the rod consists of "stiff" spring-like bonds for which atoms undergo small oscil

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Answer: hello options related to your question is missing attached below is the missing part of your question

answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )

Explanation:

When the Charge is first, Furthest away and second and closest to the source charge. <em>The spring like bonds can be said to have No charge of the length of the bonds expected because the rod did not touch the charge source </em><em>when Furthest away the bond with charge will be less effective </em>

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What is analogy a powerful rhetorical device?

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The other person who answered this is wrong btw

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A sailboat moves north for a distance of 10.00 km when blown by a wind from the exact south with a force of 5.00 x 10^4 n. how m

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(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o

IgorLugansk [536]

Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

$W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J$

So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.

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