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Ahat [919]
3 years ago
15

A group of astrologers have written a book about how to use the stars to help you choose the right friends. This could be consid

ered an example of ___________.
A. pseudoscience because it is backed by scientific studies.B. pseudoscience because it involves beliefs, not facts.C. science because we can study the stars.D. science because it is based on beliefs.
Physics
1 answer:
Mrac [35]3 years ago
8 0

Answer:b- Psuedoscience

Explanation:

It is a pseudoscience because it involves beliefs and practices which is mistakenly regarded as being based on scientific grounds.

No solid evidence is there which can prove that astrology is authentic and true that's why it is a Psuedo science

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A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan
Nesterboy [21]

Answer:

Compression distance: d \approx 0.102\,m

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy (K), in joules, at the expense of elastic potential energy (U), in joules, and overcoming work losses due to friction (W_{l}), in joules:

K + W_{l} = U (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2}  \cdot k \cdot d^{2} (2)

Where:

m - Mass of the block, in kilograms.

v - Final velocity of the block, in meters per second.

\mu - KInetic coefficient of friction, no unit.

g - Gravitational acceleration, in meters per square second.

s - Width of the rough patch, in meters.

k - Spring constant, in newtons per meter.

d - Compression distance, in meters.

If we know that m = 1.2\,kg, v = 2.3\,\frac{m}{s}, \mu = 0.44, g = 9.807\,\frac{m}{s^{2}}, s = 0.05\,m and k = 730\,\frac{N}{m}, then the compression distance of the spring is:

\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2}  \cdot k \cdot d^{2}

m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}

d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }

d \approx 0.102\,m

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Explanation:

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