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nata0808 [166]
3 years ago
6

Someone who wants to sell you a Superball claims that it will bounce to a height greater than the height from which it is droppe

d. Can this be?
Physics
1 answer:
sergeinik [125]3 years ago
7 0

Answer:

No

Explanation:

Unless there are other external forces, this will never be true. Because according to energy conservation, potential energy will be converted to kinetic energy as the ball falls down (so it loses height and gain speed). And vice versa, kinetic to potential when it bounces back. So the potential energy after must be the same (or smaller if losing heat to external environment), so it can only get the the same height or less, but not more.

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Explanation:

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The magnetic field at the center of a 1.50-cm-diameter loop is 2.70 mT . Part A. What is the current in the loop?
elixir [45]

Explanation:

It is given that,

Diameter of the circular loop, d = 1.5 cm

Radius of the circular loop, r = 0.0075 m

Magnetic field, B=2.7\ mT=2.7\times 10^{-3}\ T

(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :

B=\dfrac{\mu_o I}{2r}

I=\dfrac{2Br}{\mu_o}

I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}

I = 32.22 A

(b) The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

r=\dfrac{\mu_o I}{2\pi B}

r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}

r = 0.00238 m

r=2.38\times 10^{-3}\ m

Hence, this is the required solution.

8 0
3 years ago
A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?​
lubasha [3.4K]

At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if <em>n</em> is the number of moles of this gas, then

<em>n</em> / (19.2 L) = (1 mole) / (22.4 L)   ==>   <em>n</em> = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / <em>n</em> ≈ 14.0 g/mol

4 0
2 years ago
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