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Ugo [173]
3 years ago
13

When powdered iron is left exposed to the air, it rusts. explain why the mass of the rust is greater than the mass of the powder

ed iron?
Physics
1 answer:
Stells [14]3 years ago
7 0
There is more surface area to rust even though microscopic. which is why there is more rust than iron
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Phys-1A Horizontal Practice 2
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2 years ago
Which of Newton's Three Laws Applies? Law 1, 2, or 3?
frozen [14]

Answer:

1. Newton's first law

2.Newton's second law

3.Newton's third law

Explanation:

1. Newton's first law stated, In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force... this is base of the concept of inertia.

2. Newton's second law stated, In an inertial frame of reference, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma, or in easier words, F is directly proportional to a.

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3 years ago
Explain about kinetic theory​
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Answer:

The model, called the kinetic theory of gases, assumes that the molecules are very small relative to the distance between molecules. ... The molecules are in constant random motion, and there is an energy (mass x square of the velocity) associated with that motion. The higher the temperature, the greater the motion.

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3 years ago
Newtons laws of motion
Viktor [21]

Answer:

Law 1. A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.

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7 0
3 years ago
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

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Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

7 0
3 years ago
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