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Sergeeva-Olga [200]
4 years ago
8

How many atoms are there in 3.4moles if helium ,show the calculation

Physics
1 answer:
prohojiy [21]4 years ago
6 0
Answer: 20.4752789138x x  10^23 atoms
To count how many atoms in moles you need to know Avogadro's number. Avogadro's number dictate that for every mole there is 6.022140857 × 10^23 molecule/atoms.
Then 3.4 moles of helium will be 3.4x 6.022140857 x  10^23 atoms= 20.4752789138x x  10^23 atoms
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he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?
Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}

With the following condition :

  • \sf{\omega} = angular frequency (rad/s)
  • \sf{\theta} = change of angle value (rad)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • \sf{\theta} = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

\sf{\omega = \frac{\theta}{t}}

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0
2 years ago
A frictionless spring with a 3-kg mass can be held stretched 0.8 meters beyond its natural length by a force of 40 newtons. If t
Free_Kalibri [48]

Answer:

Explanation:

mass m = 3 kg

spring constant be k

k x .8 = 40 N

k = 40 / .8 = 50 N /m

angular frequency ω = √ ( k / m )

= √ ( 50 / 3 )

= 4.08 rad /s

Let amplitude of oscillation be A .

1/2 k A² = 1/2 m v²

50 A² = 3 x 1²

A = .245 m = 24.5 cm

For displacement , the equation of SHM is

x = A sinωt

= 24.5 sin4.08 t

x = 24.5 sin4.08 t

Here, angle 4.08 t is in radians .

3 0
3 years ago
A material through which electronics do NOT easily flow is
ziro4ka [17]
Resistor. The answer is a resistor. They do not transmit electricity. 
6 0
3 years ago
One difference between mixture and pure substances is that
Lunna [17]
Pure substances can be elements made up exclusively of one kind of atom, or they can be compounds made up of molecules that include two or more elements. Mixtures can be homogeneous or heterogeneous depending on how finely mixed the components are.
7 0
3 years ago
An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie
leva [86]

To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

MV=mv

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

V = \frac{mv}{M}

Also we have that m/M = 1/7 times

On the other hand we have from law of conservation of energy that

W_f = KE

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

F_f*S = \frac{1}{2}MV^2

\mu M*g*S = \frac{1}{2}MV^2

\mu g*S = \frac{1}{2}( \frac{mv}{M})^2

\mu = \frac{1}{2} (\frac{1}{7}v)^2

\mu = \frac{1}{98}v^2

\mu = \frac{1}{g(98)(5.1)}v^2

Here we can apply the law of conservation of energy for light mass, then

\mu mgs = \frac{1}{2} mv^2

Replacing the value of \mu

\frac{1}{g(98)(5.1)}v^2  mgs = \frac{1}{2}mv^2

Deleting constants,

s= \frac{(98*5.1)}{2}

s = 249.9m

7 0
3 years ago
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