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3 years ago

The helicopter was deformed and destroyed in the ?_ collision.

Physics

1 answer:

MariettaO [177]3 years ago

5 0

Answer:

The helicopter was deformed and destroyed in the inelastic collision.

Explanation:

When two object collide there exist two way of colliding: elastic collision and inelastic collision.
Two terms are considered during the collision: kinetic energy and momentum.
If both of these terms are conserved in any collision then there is no significant loss of property, this is called as elastic collision.

If only momentum is conserved but kinetic energy is converted into other forms then it is inelastic collision. In inelastic collision, the energy is lost in the form of vibration, sound etc. causing the damage to colliding object.

Hence the deformation of helicopter was due to inelastic collision.

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A certain metal wire has a cross sectional area of 1 mm2 and is 1 m long. when it is hung from the ceiling and a 10 kg mass is h

kvasek [131]

From the Hooke's law , the extension force of an elastic material is directly proportional to the extension.
That is, F = k e, where F is the force , k is the constant and e is the extension
F = 10 × 10 = 100 N
e = 1mm or 0.001 m
Hence, k = F/e
= 100 N/ 0.001
= 100000 N/m or 100 N/mm

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3 years ago

Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l

Masja [62]

Answer:

a) I2 = 3 (o-10) / (o- 30) , b) h ’/h= 3 (o-10) / o (o-30)

Explanation:

The builder's equation is

1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

1 / i1 = 1 / f - 1 / o

1 / i1 = 1/15 - 1 / o

i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

o2 = d-i1

We write the builder equation

1 / f2 = 1 / o2 + 1 / i2

1 / i2 = 1 / f2 -1 / o2

1 / i2 = 1 / f2 - 1 / (d-i1)

1 / i2 = 1/20 - 1 / (d-i1) (1)

Let's evaluate the last term

d-i1 = d - 15 o / (o-15)

d-i1 = (d (o-15) - 15 o) / (o-15)

d- i1 = (30 or -30 15 -15 o) / (o-15)

d-i1 = (15 or - 450) / (o- 15)

d-i1 = = (15 or -450) / (o-15)

replace in 1

1 / i2 = 1/20 - (or - 15) / (15 or -450)

1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

1 / i2 = (-5 or -150) / (15 or -150)

1 / i2 = (or -30) / (3 or - 30)

I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

m = h ’/ h = - i / o

h ’= - h i / o

In our case

h ’= h i2 / o

h ’= h 3 (o-10) / o (o-30)

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3 years ago

The helicopter was deformed and destroyed in the __?___ collision.

The helicopter was deformed and destroyed in the ?_ collision.