3 years ago

Question:

Physics

1 answer:

KengaRu [80]3 years ago

3 0

Answers:

1A) Al203

1B) SF6

2) Fe203 - iron oxide

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Using newtons second law of motion, how fast for 100 KG object accelerates 350 N of force is applied to

fenix001 [56]

Answer:

3.5m/s^2

Explanation:

From Newton's second Law of Motion

F = ma

Where F is the applied force, m is the mass of the object and a is the acceleration.

F = 350 N

Mass = 100kg

350N = 100×a

a = 350/100

a = 3.5m/s^2

The acceleration of the object will be 3.5m/s^2

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What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$