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2 years ago

Weight and mass are interchangeable terms. True or False

2 answers:

8 0

False
weight (N-force) = mass(kg-matter) x g

strictly in Physics, both are not interchangeable.
in normal life, weight (kg) actally means mass in Physics.

Nonamiya [84]2 years ago

5 0

It's false. Mass is a way of measuring how much matter an object contains, where as weight measures how hard gravity is pulling on an object. While on earth, these are typically interchangeable. However, if you were to go to Mars, your mass would stay the same, but the weight will be different. This is because you still contain the same amount of matter, but the gravity's pull will be different because the moon has a different gravitational pull than the earth. Hope this helps!

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Discuss why the article says the fact that we’re in our universe complicates our understanding of the expansion of the universe.

MArishka [77]

<span>Because of our perception of the universe from inside the universe, we are unable to see how and towards what the universe is expanding. Also, our understanding of it is further complicated because we are moving as part of the expansion, thus distorting our perception of it.</span>

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2 years ago

Read 2 more answers

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

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3 years ago

If the temperature of an iron sphere is increased A)its density will decrease .B)its volume will decrease. C)its mass will decre

babymother [125]

Answer:

A) Its density will decrease

Explanation:

When an object is heated, its volume increases. This is due to the fact that the particles in the medium vibrate more (if it is a solid) or they move more (if it is a liquid or a gas), therefore they tend to occupy a larger space.

At the same time, the mass of the object does not change, because the mass just represents the amount of matter contained in the object, so it does not increase/decrease at different temperatures.

The density of an object is defined as the ratio between the mass (m) and the volume (V):

$d=\frac{m}{V}$

We said that the mass remains unchanged while the volume increases: since the density is inversely proportional to the volume, this means that the density decreases.

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2 years ago

Read 2 more answers

how much chemical energy must be supplied to a car engine if we want it to produce 5000J of kinetic energy and it is 45% efficie

vagabundo [1.1K]

Here i state the conservation of energy rule and use that to justify my answer. I showed how to manipulate percentages to get the final answer of 11000J (2sf). Hope I'm right xx