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3 years ago

The helicopter was deformed and destroyed in the _______ collision.

2 answers:

8 0

The type of collision shown is an inelastic collision. Collisions happen when two bodies in motion come in contact with each other. Inelastic collision are experienced when the momentum of each body is not conserved, therefore the bodies come to a complete stop.

ASHA 777 [7]3 years ago

7 0

There are two types of collision in physics which are named as elastic collision and inelastic collision.

Elastic collision is that type of collision in which both kinetic energy and momentum is conserved i.e total kinetic energy and momentum before collision is equal to the total kinetic energy and momentum after collision.As kinetic energy is conserved here,so less or ideally no damage occurs to the colliding bodies.

Inelastic collision is that type of collision in which only momentum is conserved but not kinetic energy .After the collision the kinetic energy is lost in forms heat,sound ,light etc.

Normally any deformation and destruction is due to the inelastic collision which results in the loss of kinetic energy.

Hence the helicopters were destroyed and deformed due to inelastic collision.

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Vector C has a magnitude of 24.6 m and points in the − y ‑ direction. Vectors A and B both have positive y ‑ components, and mak

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Answer:

A= 61.35

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Explanation:

1. Using the components method we have:

$A_{x}= A cos \alpha\\B_{x}= B cos \beta\\C_{x}= 0\\\\A_{y}= A sin \alpha\\B_{y}= B sin \beta\\C_{y}= 24.6\\$

Considering that the vector sum $A+B+C=0$ , then:

$|V|=\sqrt{V_{x}^{2} +V_{y}^{2} }=0$

Then:

$V_{x} ^{2} =0; V_{x} =0\\V_{y} ^{2} =0; V_{y} =0$

It means the value of x and y component is 0.

2. Determinate the equations that describe each component:

$V_{x}= A cos \alpha -B cos \beta=0 (1)\\V_{y}= A sin \alpha +B sin \beta - C=0 (2)$

Form Eq. (1):

$A=B \frac{cos \beta}{cos \alpha} (3)$

Replacing A in Eq. (2):

$(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-C=0\\(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-=C\\\\B(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)=C\\B=C(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)^{-1} (4)$

Replacing values of C, α and β in (4):

$B= 24.6 (\frac{(cos 27.7)(sin 44.9)}{cos 44.9}+sin 27.7)^{-1} \\B= -44.4$

Replacing value of B in (3)

$A=-44.40\frac{cos 27.7}{sin 49.9} \\A= 61.35$

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Answer:

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