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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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Which of the following is the accepted scientific opinion about the current warming trend?
bija089 [108]
The answer is there is no one accepted scientific opinion about the current warming trend in the global climate 
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When will you say a body is in a) uniform acceleration (b) non uniform acceleration
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Answer:

See below ~

Explanation:

Part (a) :

We can say a body is in uniform acceleration if the acceleration of the object remains constant with respect to time throughout its motion.

Part (b) :

We can say a body is non-uniform acceleration if the acceleration of the body varies with respect to time throughout its motion.

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2 years ago
(BRAINLIEST)
tankabanditka [31]
  • Cm=C
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A swing that is 9.2m long has a child in it who is swinging. what is its period in seconds
Fantom [35]
The child on a swing can be modeled as a simple pendulum.  The period of a simple pendulum is given by

T=2 \pi \sqrt{ \frac{L}{g} }=2 \pi \sqrt{ \frac{9.2m}{9.8 \frac{m}{s^2} } }=6.09s
8 0
3 years ago
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle res
zhannawk [14.2K]

Answer:

a) 250 N/m

b) 22.4 rad/s , 3.6 Hz , 0.28 sec

c) 0.3125 J

Explanation:

a)

F = force applied on the spring = 7.50 N

x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m

k = spring constant of the spring

Since the force applied causes the spring to stretch

F = k x

7.50 = k (0.03)

k = 250 N/m

b)

m = mass of the particle attached to the spring = 0.500 kg

Angular frequency of motion is given as

w = \sqrt{\frac{k}{m}}

w = \sqrt{\frac{250}{0.5}}

w = 22.4 rad/s

f = frequency

Angular frequency is also given as

w = 2 π f

22.4 = 2 (3.14) f

f  = 3.6 Hz

T = Time period

Time period is given as

T = \frac{1}{f}

T = \frac{1}{3.6}

T = 0.28 sec

c)

A = amplitude of motion = 5 cm = 0.05 m

Total energy of the spring-block system is given as

U = (0.5) k A²

U = (0.5) (250) (0.05)²

U = 0.3125 J

5 0
4 years ago
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