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Elena L [17]
2 years ago
7

Need help with the following - question 2

Physics
1 answer:
Leona [35]2 years ago
6 0

Answer:

Find the change in momentum of the upper stage, that is:

∆p = m(vf - vi)

m being the mass of the upper stage

vf being the final velocity which was given

vi being the initial which was also given

find ∆p

then use ∆p in the same equation

∆p being the answer you got above

m being the mass of the lower stage (given)

vi being the initial velocity (given)

vf being the final velocity of the lower stage which you were asked to find

Explanation:

During a collision the change in momentum (∆p) for both objects is equal regardless of their speeds or masses before or after the equation

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The bar graph shows energy data taken from a roller coaster at a theme park. analyze the data and assess its validity. 3-5 sente
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This question involves the concepts of the law of conservation of energy, potential energy, and kinetic energy.

The data shown by the bar graph is "valid".

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Considering the bottom of the first hill:

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From the data given in the bar graph:

Total Energy at 1st Hill = 2,500,000 J + 0 J (since at the bottom potential energy is zero due to zero height)

Total Energy at 1st Hill = 2,500,000 J

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Total Energy at 2nd Hill = Kinetic Energy + Potential Energy

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Total Energy at 2nd Hill = 2,500,000 J

Hence,

Total Energy at 1st Hill = Total Energy at 2nd Hill

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Learn more about the law of conservation of energy here:

brainly.com/question/20971995?referrer=searchResults

The attached picture explains the law of conservation of energy.

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A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

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Divide both sides by L to obtain;

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Thus;

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K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

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3 years ago
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