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lesya [120]
2 years ago
11

In a study of momentum during collisions, a student threw an egg at a large sheet of cloth that was suspended from a line and ob

served that the egg did not break during the collision. A second egg, thrown with similar force at a suspended concrete block, did break on impact. Why did the first egg survive the collision while the second egg did not survive?.
Physics
1 answer:
Setler [38]2 years ago
6 0

From the law of conservation of momentum, the first egg did not break because it forward momentum is transferred to the cloth while the second egg broke, the block exerts an opposite force equal to the momentum of the egg on the egg.

<h3>Momentum</h3>

Momentum is the product of the mass of an object and its velocity of motion.

The law of conservation of momentum states that the momentum of an isolated system of colliding bodies is conserved.

When an egg is thrown on a large sheet of cloth suspended, the egg does not break because the momentum of the egg is transferred to the cloth as it is depressed.

However, a second egg, thrown with similar force at a suspended concrete block, did break on impact because the block does not move forward, instead it exerts an equal but opposite force on the egg resulting in the egg breaking.

Therefore, from the law of conservation of momentum, the first egg did not break because it forward momentum is transferred to the cloth while the second egg broke, the block exerts an opposite force equal to the momentum of the egg on the egg.

Learn more about momentum at: brainly.com/question/7538238

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4.689 years

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If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh
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Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

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            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

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3 years ago
The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at
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Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

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A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

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but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

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Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

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C)

using eqn (2)

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m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

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