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Nutka1998 [239]
3 years ago
10

How are some types of collisions different from others?

Physics
1 answer:
lyudmila [28]3 years ago
3 0
There are two general types of collisions, inelastic and elastic. 
Inelastic collisions occur when two objects collide but neither of them bounce away from each other.
Collisions in which the objects do not touch each other are elastic. (Ex: Rutherford Scattering) 
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If the star Alpha Centauri were moved to a distance 10 times farther than it is now, its parallax angle would
7nadin3 [17]

Answer:

decrease by a factor 10

Explanation:

The parallax angle of a close star is given by

p=\frac{1}{d}

where

p is the parallax angle

d is the distance of the star from Earth, in parsecs

From the formula we see that the parallax angle is inversely proportional to the distance.

In this problem, the distance of the star is increased by a factor 10:

d' = 10 d

so the new parallax angle would be

p'=\frac{1}{10 d}=\frac{1}{10}\frac{1}{d}=\frac{p}{10}

So, the parallax angle would decrease by a factor 10.

4 0
3 years ago
Calcular la rapidez promedio
Vlada [557]

Answer:

  v_average = 15 m / s

Explanation:

The average speed can be found in two ways,

* taking the distance traveled and divide it by the time spent

* taking the velocities in each time interval and then finding the weighted average by the time fraction

              v_average = 1 / t_total ∑  v_{i}  t_{i}vi ti

Let's apply this last equation

               

Total time is

               t = t₁ + t₂

               t = 10 + 10 = 20   min

              v_average = 10/20 10 + 10/20 20

              v_average = 10/2 + 20/2

              v_average = 15 m / s

7 0
4 years ago
Estimate the change in the equilibrium melting point of copper caused by a change in pressure of 10 kbar. The molar volume of co
mr Goodwill [35]

Answer:

The change in the equilibrium melting point is 4.162 K.

Explanation:

Given that,

Pressure = 10 kbar

Molar volume of copperV=8.0\times10^{-6}\ m^3

Volume of liquid V=7.6\times10^{-6}\ m^3

Latent heat of fusion L= 13.05 kJ

Melting point =1085°C

We need to calculate the change temperature

Using Clapeyron equation

\dfrac{\Delta P}{\Delta T}=\dfrac{\Delta H}{T\Delta V}

Put the value into the formula

\dfrac{1000\times10^{5}}{\Delta T}=\dfrac{13050}{(1085+273)\times(8.0-7.6)\times10^{-6}}

\Delta T=\dfrac{1000\times10^{-5}\times(1085+273)\times(8.0-7.6)\times10^{-6}}{13050}

\Delta T=4.162\ K

Hence, The change in the equilibrium melting point is 4.162 K.

5 0
3 years ago
What grade of sprain is a completely torn ligament?
sattari [20]

Knee sprains can be classified into three grades, depending on the amount of damage: Grade 1: a few fibres (less than 10%) are damaged/torn. Usually heals naturally. Grade 2: more fibres are torn but the ligament is still intact. Grade 3: the ligament is ruptured ie completely torn.

:)

5 0
3 years ago
Read 2 more answers
A cylindrical container with a cross-sectional area of 85.2 cm 2 holds fluid having a density of 806 kg/m 3 . The surface of the
Anton [14]

Answer:

the answer would be B your welcome

Explanation:

5 0
3 years ago
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