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3 years ago

Describe another real-world example of a perfectly inelastic collision

Physics

2 answers:

Sidana [21]3 years ago

8 0

1. Newtons Cradle
2. When you play pool and one ball transfers all its momentum, making the first ball come to a halt.

Ivahew [28]3 years ago

5 0

An example of a perfectly elastic collision is that it doesn't exist. But for the sake of saying it exist you would assume that a collision takes place where energy is completely conserved and is not lost as (non-usable energy) heat. Maybe you say that a gas molecule collides with a completely (perfectly) smooth surface and the gas molecules collides perpendicular to the surface (does not collide at an angle). The gas molecule then bounces off the wall with the same speed as when it colliding against the wall. This would assume a perfectly elastic collision as KE=1/2mv^2. When your gas molecule collides its mass doesn't change but when you say that the gas molecule had the same velocity when it bounces off the wall you ahve assumed that kinetic energy is maintained (which assumes perfectly elastic).

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A weight suspended from a spring is seen to bob up and down over a total distance of 20 centimeters twice each second.

Tom [10]

"Twice each second" means "2 per second". Frequency = 2 Hz.

Period = time for one bounce = 1 second

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Amplitude = 10 centimeters

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2 years ago

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3 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

8 0

3 years ago

What layers on earth are considered liquids

Tatiana [17]

Crust. The thin solid outermost layer of Earth. ...Asthenosphere. The lower layer of the crust. ...Lithosphere.Plasticity: is solid but still being able to. flow without being a liquid.The cool, rigid outermost layer of the Earth. ...the solid part of the earth consisting of the crust and outer mantle.

6 0

3 years ago

what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?

nignag [31]

Using Newton's second law of motion:

F=ma ; [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)

Given: Find: Formula: Solve for m:

F: 2500N mass:? F=ma Eq.1 m=F/a Eq. 2

a= 200m/s^2

Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

8 0

3 years ago