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4 years ago

What is an elastic collision?

Physics

1 answer:

Anit [1.1K]4 years ago

6 0

Answer:

A collision in which both total momentum and total kinetic energy are conserved

Explanation:

In classical physics, we have two types of collisions:

- Elastic collision: elastic collision is a collision in which both the total momentum of the objects involved and the total kinetic energy of the objects involved are conserved

- Inelastic collision: in an inelastic collision, the total momentum of the objects involved is conserved, while the total kinetic energy is not. In this type of collisions, part of the total kinetic energy is converted into heat or other forms of energy due to the presence of frictional forces. When the objects stick together after the collision, the collisions is called 'perfectly inelastic collision'

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A light ray travels in the +x direction and strikes a slanted surface with an angle of 62° between its normal and the ty axis. T

Elena-2011 [213]

Answer:

- 0.6

Explanation:

Given that angle between normal y axis is 62° so angle between normal

and x axis will be 90- 62 = 28 °. Since incident ray is along x axis , 28 ° will be the angle between incident ray and normal ie it will be angle of incidence

Angle of incidence = 28 °

angle of reflection = 28°

Angle between incident ray and reflected ray = 28 + 28 = 56 °

Angle between x axis and reflected ray = 56 °

x component of reflected ray

= - cos 56 ( it will be towards - ve x axis. )

- 0.6

6 0

3 years ago

20 pts !!!!

garri49 [273]

Am not really sure but what i see is D

4 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

State the name of the process in each of the following changes of state:

polet [3.4K]

Answer:

A. Melting

B. Freezing

C. Condensing

D. Boiling

Explanation:

4 0

4 years ago

If a girl is standing still and holding a box, is she doing any work? Why or why not?

sukhopar [10]

I beleive she isnt doing any work due to holding the box motionless, you must be exerting a force in the direction of the box motion. If she is just standing there holding the box their isn't no work becuase no distance has been covered. work = force = distance.

7 0

3 years ago

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