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3 years ago

6

PLZ HELP: Which food items would be appropriate for a spacecraft? Check all that apply. A< brownie, B< dry salt and pepper

, C

1 answer:

Sladkaya [172]3 years ago

7 0

B. where is the rest of C?
My answer is B

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Name the type of particle formed by covalent bonds

jasenka [17]

Molecules and polyatomic ions are formed by covalent bonds.

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2 years ago

As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over th

Fofino [41]

Answer:

$5.04\cdot 10^8 A$

Explanation:

The work function of the metal corresponds to the minimum energy needed to extract a photoelectron from the metal. In this case, it is:

$\phi = 3.950\cdot 10^{-19}J$

So, the energy of the incoming photon hitting on the metal must be at least equal to this value.

The energy of a photon is given by

$E=\frac{hc}{\lambda}$

where

h is the Planck's constant

c is the speed of light

$\lambda$ is the wavelength of the photon

Using $E=\phi$ and solving for $\lambda$ , we find the maximum wavelength of the radiation that will eject electrons from the metal:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.950\cdot 10^{-19} J}=5.04\cdot 10^{-7}m$

And since

1 angstrom = $10^{-15}m$

The wavelength in angstroms is

$\lambda=\frac{5.04\cdot 10^{-7} m}{10^{-15} m/A}=5.04\cdot 10^8 A$

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3 years ago

1. Which of the following is not included

Harrizon [31]

Answer:

cartilage

Explanation:

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2 years ago

One nanometer is equal to how many centimeters?

Neko [114]

One nanometer = 1e-7

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2 years ago

Read 2 more answers

A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

bixtya [17]

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$

Substitute the given values

So

$u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56$

6 0

3 years ago