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prisoha [69]
3 years ago
5

If two billiard balls collide, what happens to the systems overall momentum ?

Physics
1 answer:
andreyandreev [35.5K]3 years ago
4 0
Remains same. Here according to law of conservation of momentum, the system have momentum conserved and remains constant.
Given by the popular equation,
m1v1=m2v3
m1=Mass of billiard ball 1
m2= Mass of the 2nd ball
v1,v2 are the velocity of the ball 1,2 respectively.
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Dimension of radius of sphere​
Readme [11.4K]

Answer:

The dimension is L

Explanation:

Dimension analysis is a method of representing quantities majorly with respect to some fundamental quantities of mass (M), length (L), time (T).

A sphere has a definite volume which relates to its radius by:

V = \frac{4}{3}\pir^{3}

In this equation \pi is a dimensionless quantity, and the unit of v is m^{3}.

But, metre is a measure of length, thus it has a dimension of L.

So that,

m^{3} ≅ L^{3}

Then,

L^{3} = r^{3}

Find the cube root of both sides to have,

r = L

Therefore, the dimension of the radius of a sphere is L.

5 0
3 years ago
What is the density of iron if 5.0 cm3 has a mass of 39.5g
weqwewe [10]

Answer :\rho = 7.9\ g/cm^3

Explanation :

It is given that

Mass of iron, m = 39.5 g

Volume of iron, V = 5\ cm^3

So, density is :

density, \rho =\dfrac{mass}{volume}

\rho =\dfrac{39.5g}{5cm^3}

\rho = 7.9\ g/cm^3

7 0
3 years ago
Which year was pluto no longer considered a planet?.
katrin2010 [14]
2006, i hope this helps
6 0
2 years ago
The thunderbolt bobsled team is training for Olympic Gold. During practice they start a run with a speed of 0.57 m/s, they compl
aleksandr82 [10.1K]
acceleration=\frac{\Delta\ velocity}{\Delta\ time}\\\\
v_{initial}=0,57m/s\\
distance=1360m\\ \Delta\ time=89,49seconds\\\\
v_{final}-v_{initial}=\frac{distance}{time}\\
v_{final}=\frac{distance}{time}+v_{initial}\\
v_{final}=\frac{1360}{89,49}+0,57\\\\v_{final}=15,77\frac{m}{s}\\\\
acceleration=\frac{15,77-0,57}{89,49}=0,17\frac{m}{s^2}\\\\ \boxed{acceleration=0,17\frac{m}{s^2}}
6 0
3 years ago
Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many
Zielflug [23.3K]

Answer:

The 6.25 \times 10^{23} electrons are moving through the superconductor per second.

Explanation:

Given :

Current I = 1 \times 10^{5} A

Charge of electron e = 1.6 \times 10^{-19} C

Time t = 1 sec

From the formula of current,

Current is the number of charges flowing per unit time.

   I = \frac{ne}{t}

Where n = number of charges means in our case number of electrons

   n = \frac{It}{e}

   n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }

   n = 6.25 \times 10^{23}

Therefore, 6.25 \times 10^{23} electrons are moving through the superconductor per second.

5 0
3 years ago
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