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AlladinOne [14]
3 years ago
15

Estimate the number of atoms in 1 cm^3 of a solid

Physics
1 answer:
dexar [7]3 years ago
7 0
Avogadro's number: 6.02 x 10^23 atoms is present in 1mol of a solid (i.e. 22, 400 cm3)

Hence, in 1 cm3, 6.02 x 10^23 /22400 atoms is present = 2 x 10 ^ 19 atoms.
You might be interested in
Why is work and potential energy the same thing
kipiarov [429]
Potential energy and work done are the same thing as much as kinetic energy and work done are the same thing. Potential energy is a state of the system, a way of storing energy as of virtue of its configuration or motion, while work done in most cases is a way of chaning this energy from one body to another.
5 0
3 years ago
• How much work is<br>required to lift a 2kg<br>object 2m high?<br>​
pychu [463]

Answer You need to consider that the gravity on earth is 9.8 m/s/s. This means any object you let go on the earths surface will gain 9.8 m/s of speed every second. You need to apply a force on the object in the opposite direction to avoid this acceleration. If you are pushing something up at a constant speed, you are just resisting earths acceleration. The more massive and object is, the greater force is needed to accelerate it. The equation is Force = mass*acceleration. So for a 2kg object in a 9.8 m/s/s gravity you need 2kg*9.8m/s/s = 19.6 Newtons to counteract gravity. Work or energy = force * distance. So to push with 19.6 N over a distance of 2 meters = 19.6 N*2 m = 39.2 Joules of energy. There is an equation that puts together those two equations I just used and it is E = mgh

The amount of Energy to lift an object is (mass) * (acceleration due to gravity) * (height)

:Hence, the Work done to life the mass of 2 kg to a height of 10 m is 196 J. Hope it helps❤️❤️❤️

Explanation:

7 0
3 years ago
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

      $=\sqrt{2 \times 9.8 \times 0.35}$      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

      $=\frac{0.175}{0.525}+0$

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

 

3 0
3 years ago
A spaceship whose rest length is 350m has a speed of .82c
igomit [66]

Answer:

t'=1.1897*10^{-6} s

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}

where:

v is the velocity of spaceship relative to certain frame of reference =  -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }

u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

t'=\frac{length}{Relatie seed (u')}

t'=\frac{350}{0.9806c}     (c = 3*10^8 m/s)

t'=\frac{350}{0.9806* 3.0*10^{8} }

t'=1.1897*10^{-6} s

t'=1.1897 μs

4 0
3 years ago
A ball is thrown into the air with an upward velocity of 20 feet per second. Its height, h, in feet after t seconds is given by
Margarita [4]
Thank you for posting your question here at brainly. Feel free to ask more questions.  
<span>The best and most correct answer among the choices provided by the question is B. Reaches a max height of 8.25 feet after 0.63 seconds</span> .     <span><span>

</span><span>Hope my answer would be a great help for you. </span> </span>  

<span> </span>

3 0
3 years ago
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