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2 years ago

Which is most likely an example of pseudoscience?

Physics

2 answers:

Montano1993 [528]2 years ago

7 0

I would say its the last answer

Leya [2.2K]2 years ago

5 0

The correct answer is option D, Most people believe that drinking milk can soothe heartburn.

Reason -

Pseudoscience is a science in which facts and belief are considered to be existing but they lack any scientific logical backing.

Here all the three options A, B and C have been proved and well established except the last one. It’s a general belief of people that drinking milk soothes out heartburn but no scientific investigation has yet accepted the fact.

You might be interested in

What advantage do space telescopes have over telescopes used on earth?

Elina [12.6K]

Earth's atmosphere blocks many types of light including gamma, x-rays most ultraviolet and infrared. So optical telescopes that use visible light and ultraviolet telescopes that are used to study very hot stars are much less effective on Earth.

4 0

2 years ago

Read 2 more answers

A rectangular block of copper metal weighs 1896 g. The dimensions of the block are 8.4 cm by 5.5 cm by 4.6 cm. From this data, w

sp2606 [1]

Answer:

8.9 g/cm^3

Explanation:

density = mass/volume

volume = length * width * height

volume = (8.4 cm)(5.5 cm)(4.6 cm)

volume = 212.52 cm^3

mass = 1896 g

density = (1896 g)/(212.52 cm^3)

density = 8.9 g/cm^3

3 0

2 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

2 years ago

What is the purpose of experimentation in a research study

Nat2105 [25]

Hello,

The answer is to "prove your hypothesis".

Reason:

Researchers do experiments to prove there hypothesis they will most likely do the experiment a few times in older to have the conclusion valid therefore proving his or her experiment.

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit

6 0

2 years ago

Read 2 more answers

The mass of jupiter is 300 times the mass of the earth. Jupiter orbits the sun with Tjupiter = 11.9 yr in an orbit with Rjupiter

mihalych1998 [28]

Answer:

c) 11.9 yr

Explanation:

The orbital period is proportional to r^(3/2) and does not depend on the satellite's mass. Any object at Jupiter position will have the same orbital period regardless of mass.

By keppler's law we know that

T^2= r^3

T= orbital time period

r= mean distance of the planet from the Sun.

clearly, The orbital period does not depend on the satellite's mass

there, the correct answer will be c= 11.9 yr.

5 0

2 years ago