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3 years ago

What is the general defination of parallax?

Physics

1 answer:

Tcecarenko [31]3 years ago

3 0

The effect whereby the position or direction of an object appears to differ when viewed from different positions

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A 15.0-kg object sitting at rest is struck elastically in a head-on collision with a 10.5-kg object initially moving at 3.0 m/s.

DIA [1.3K]

Answer:

The final velocity of the 15.0-kg object after the collision is 2.47 m/s in forward direction.

Explanation:

Given;

mass of the object, m₁ = 15 kg

initial velocity of this object, u₁ = 0

mass of the second object, m₂ = 10.5 kg

initial velocity of this object, u₂ = 3.0 m/s

let the final velocity of the first object = v₁

also, let the final velocity of the second object = v₂

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(15 x 0) + (10.5 x 3) = 15v₁ + 10.5v₂

31.5 = 15v₁ + 10.5v₂ ----- (1)

One directional velocity;

u₁ + v₁ = u₂ + v₂

0 + v₁ = 3 + v₂

v₂ = v₁ - 3 ------(2)

Substitute (2) into (1);

31.5 = 15v₁ + 10.5v₂

31.5 = 15v₁ + 10.5(v₁ - 3)

31.5 = 15v₁ + 10.5v₁ - 31.5

63 = 25.5v₁

v₁ = 63 / 25.5

v₁ = 2.47 m/s

Therefore, the final velocity of the 15.0-kg object after the collision is 2.47 m/s in forward direction.

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3 years ago

A crate falling from a plane is part of a closed system that has 24,450 J of

Inessa05 [86]

It falls down and I hate that this makes me type a lot

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3 years ago

How many valence electrons does each atom of arsenic (As) have? Arsenic is element 33. It is in period 4 and family 15 (5A or th

vodka [1.7K]

Answer:

It have 5 valence electrons

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3 years ago

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The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t

maxonik [38]

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

The work done by the resistance force.

Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s$

Work done is equal to the change in kinetic energy. It can be given as follows :

$W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J$

So, the required work done is 32.99 J.

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3 years ago

What is missing from this graph?<br><br> (A) A title.<br> (B) Axis labels.<br> (C) A trendline.

SIZIF [17.4K]

Axis labels. We don’t know what is being compared with the data

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3 years ago

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