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2 years ago

Why is an object’s mass, rather than its weight, used to indicate the amount of matter it contains?.

1 answer:

3 0

An object's mass, rather than its weight is used to indicate the amount of matter it contains because weight is defined as the amount of force due to an existing field (In most cases, gravitational field) that is experienced by the body. The weight of the same body can be different in different environments it's observed.

However, the mass of the body is the exact measure of the amount of matter contained in a body. Which is constant, regardless of the environment, or conditions the body is observed.

Hence, for real-world calculations and experimentation, the mass of the body is considered to represent the amount of matter it contains rather than its weight.

Refer more about mass here

brainly.com/question/14651380

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A simple harmonic oscillator of amplitude A has a total energy E.

jeka57 [31]

Answer:

a) $K=E-\frac{kA^2}{18}$

b) $U=\frac{kA^2}{18}$

Explanation:

The law of conservation of mechanical energy states that total mechanical energy remains constant during oscillation. Mechanical energy is defined as the sum of kinetic energy and potential energy:

$E=U+K\\E=\frac{kx^2}{2}+\frac{mv^2}{2}$

a) The position is one-third the amplitude. So, we have $x=\frac{1}{3}A$ . Replacing and solving for K

$E=\frac{k(\frac{1}{3}A)^2}{2}+K\\E=\frac{kA^2}{18}+K\\K=E-\frac{kA^2}{18}$

b) The potential energy is defined as:

$U=\frac{kx^2}{2}$

Replacing:

$U=\frac{k(\frac{1}{3}A)^2}{2}\\U=\frac{kA^2}{18}$

4 0

3 years ago

Velocity differs from speed in that velocity indicates a particle's __________ of motion.

masha68 [24]

Velocity differs from speed in that velocity indicates a particle's direction of motion.

Therefore, your correct answer is: Direction

Good luck with your studies, I hope this helps~!

5 0

3 years ago

A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.

Nataliya [291]

Answer:

Approximately $1.2 \times 10^{4}\; {\rm J}$ (assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The change in the gravitational potential energy ( ${\rm GPE}$ ) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass $m$ increased by $\Delta h$ , the ${\rm GPE}$ of that object would have increased by $m\, g\, \Delta h$ .

In this question, the height of this object increased by $\Delta h = 6.0\; {\rm m}$ . The mass of this object is $m = 200\; {\rm kg}$ . Thus, the ${\rm GPE}$ of this object would have increased by:

$\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}$ .