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wel
3 years ago
13

A gold bar contains 63.0 mol of gold, Au(s). How many atoms of gold are in the bar?

Physics
1 answer:
vredina [299]3 years ago
7 0

Answer:

63.0 x (6.02 x 10^23) = 3.79 x 10^25 atoms of gold

Explanation:

There are 6.02 x 10^23 representative particles in a mole of any substance. Multiply it by the number of moles, and you will get the number of representative particles (atoms, molecules, formula units)

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What is the kinetic energy of a 5 kg object moving at 4 m/s
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Answer:

40

Explanation:

rule: K.E = 1/2 m×v^2

1/2 × 5 × (4)^2= 40 joule

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3 years ago
A Describe and explain what
melomori [17]

Answer:

The extension is directly proportional to the force applied.

ex: if the force is doubled, the extension doubles. This works until the limit of proportionality is exceeded.

Hope this helped~

Explanation:

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2 years ago
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A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matt
anyanavicka [17]

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

6 0
3 years ago
Write down the symbols of nitrogen and neon?​
statuscvo [17]

Answer:

Nitrogen is N, neon is Ne

5 0
3 years ago
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A cello string vibrates in its fundamental mode with a frequency of 335 1/s. The vibrating segment is 28.5 cm long and has a mas
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Answer:

The tension in string is found to be 188.06 N

Explanation:

For the vibrating string the fundamental frequency is given as:

f1 = v/2L

where,

f1 = fundamental frequency = 335 Hz

v = speed of wave

L = length of string = 28.5 cm = 0.285 m

Therefore,

v = f1 2L

v = (335 Hz)(2)(0.285)

v = 190.95 m/s

Now, for the tension:

v = √T/μ

v² = T/μ

T = v² μ

where,

T = Tension

v = speed = 190.95 m/s

μ = linear mass density of string = mass/L = 0.00147 kg/0.285 m = 5.15 x 10^-3 kg/m

Therefore,

T = (190.95 m/s)²(5.15 x 10^-3 kg/m)

<u>T = 188.06 N</u>

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3 years ago
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