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4 years ago

Describe another real-world example of a perfectly inelastic collision

Physics

2 answers:

Sidana [21]4 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]4 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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Ano -ano ang uri ng terorismo

kogti [31]

Answer:

nagaganap sa loob ng bansa sa pagitan ng mga pinuno

6 0

4 years ago

Suppose we take a 1 m long uniform bar and support it at the 21 cm mark. Hanging a 0.40 kg mass on the short end of the beam res

Ipatiy [6.2K]

Answer:

The mass of the beam is = 29 kg.

Explanation:

A beam with mass 40 kg is shown in figure. Point S is the support point. Point B is the middle point on the beam where mass of the beam acts.

Taking moment about Point S

40 × 21 = $M_{beam}$ × 29

$M_{beam}$ = 29 kg

Therefore the mass of the beam is = 29 kg.

4 0

3 years ago

Tasha has mass 20 kg and wants to use a 4.0-m board of mass 10 kg as a seesaw. her friends are busy, so tasha seesaws by herself

ELEN [110]

The 4 m board's center of mass is 2 m so the pivot point is somewhere between Tasha and 2 m (if you draw a picture it's going to make this clearer)

The mass of Tasha * her distance from the pivot point = the board's mass * distance from the pivot point.
 Let d1 be the distance Tasha is from the pivot point. The board's center of mass from the pivot point is 2 m - d1
 20 kg * d1 = 10 kg * (2m - d1)
 Solve for d1, you should get 2/3 of a m for the distance Tasha is from the pivot or support point. The center of mass for the board is 1 1/3 m from the support point.

8 0

3 years ago

A cup of hot coffee initially at 95 degrees C cools to 80 degrees C in 5 min while sitting in a room of temperature 21 degrees C

Oksana_A [137]

Answer:

When the temperature of the coffee is 50 °C, the time will be 20.68 mins

Explanation:

Given;

The initial temperature of the coffee T₀ = 95 °C

The temperature of the room = 21°C

Let T be the temperature at time of cooling t in mins

According to Newton's law of cooling;

$\frac{dT}{dt} \alpha (T-21)\\\\\frac{dT}{dt} = k (T-21)\\\\\frac{dT}{T-21} = kdt\\\\\int\limits {\frac{dT}{T-21}} = \int\limits kdt\\\\Log(T-21) =kt + Logc \\\\Log (\frac{T-21}{c} ) = kt\\\\T -21 = ce^{kt}\\\\At \ t = 0, T = 95\\\\95-21 = ce^0\\\\74 = c\\\\New, equation: T -21 = 74e^{kt}\\\\Again; when \ t= 5\ min, T = 80\\\\80 -21 = 74e^{5k}\\\\59 = 74e^{5k}\\\\e^{5k} = \frac{59}{74}\\\\ 5k = ln(\frac{59}{74})\\\\5k = -0.2265\\\\k = -0.0453$

When the temperature is 50 °C, the time t in min is calculated as;

$T -21 = 74e^{-0.0453t}\\\\50 -21 = 74e^{-0.0453t}\\\\29 = 74e^{-0.0453t}\\\\\frac{29}{74} = e^{-0.0453t}\\\\0.39189 = e^{-0.0453t}\\\\ln(0.39189 ) = {-0.0453t}\\\\-0.93677 = {-0.0453t}\\\\t = \frac{-0.93677}{-0.0453}\\\\ t = 20.68 \ mins$

Therefore, when the temperature of the coffee is 50 °C, the time will be 20.68 mins

4 0

3 years ago

Help please (about projectile motion at an angle)

PolarNik [594]

Answer:

Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao

Explanation:

7 0

3 years ago