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lys-0071 [83]
1 year ago
6

What is one advantage of using primary sources when doing research on an

Physics
1 answer:
WITCHER [35]1 year ago
5 0

One advantage of using primary sources when doing research on an experiment is, it contain more details about the experiment. Thus, option D is right .

To find the answer, we have to know more about the Primary sources.

<h3>What is the advantages of primary sources?</h3>
  • Primary sources are the direct or first-hand excerpts that scientists have independently written based on their experiments in order to support their research.
  • Utilizing original sources encourages analytical and critical thought in relation to the research.
  • It aids in exploring from various angles, which leads to the discovery of extra facts.
  • Primary data aids in navigating the conflicts. Since it is a direct resource related to the experiment, it serves as evidence for the data.

Thus, we can conclude that, Primary sources contain more details about the experiment.

Learn more about the Primary sources here:

brainly.com/question/24190082

#SPJ9

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So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th
tia_tia [17]

Answer:

a) a=33.73mm/s^{2}

b) mg>N

c) \%_{change}=0.343\%

d) a=24.07mm/s^{2}

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

a_{c}=\omega ^{2}r

where:

\omega=\frac{2\pi}{T}

we know the period of rotation of the earth is about 24 hours, so:

T=24hr*\frac{3600s}{1hr}=86400s

so we can now find the angular speed:

\omega=\frac{2\pi}{86400s}

\omega=72.72x10^{-6} rad/s^{2}

So the centripetal acceleration will be:

a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)

which yields:

a_{c}=33.73mm/s^{2}

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

\sum F=0

so we get that:

N-mg+ma_{c} = 0

and solve for the normal force:

N=mg-ma_{c}

In this case, we can clearly see that:

mg>mg-ma_{c}

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

mg=(60kg)(9.81m/s^{2})=588.6N

and let's calculate the normal force:

N=m(g-a_{c})

N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})

N=586.58N

so now we can calculate the percentage change:

\%_{change} = \frac{mg-N}{mg}x100\%

so we get:

\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%

\%_{change}=0.343\%

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

cos \theta = \frac{AS}{h}

In this case:

cos \theta = \frac{r}{R_{E}}

so we can solve for r, so we get:

r= R_{E}cos \theta

in this case we'll use the average radius of earch which is 6,371 km, so we get:

r = (6371x10^{3}m)cos (44.4^{o})

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

a=\omega ^{2}r

a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m

a=24.07 mm/s^{2}

5 0
3 years ago
Consider a current carrying a wire coming out of your computer screen towards you. Which statement below correctly describes the
Ugo [173]

Answer:

1. The magnetic field encircles the wire in a counterclockwise direction

Explanation:

When we have a current carrying wire perpendicular to the screen in which the current flows out of the screen then by the Maxwell's right-hand thumb rule we place the thumb of our right hand in the direction of the current and curl the remaining fingers around the wire, these curled fingers denote the direction of the magnetic field which is in the counter-clock wise direction.

Ever current carrying conductor produces a magnetic field around it.

5 0
3 years ago
A6 kg mass moving at 10m/s collides with a 4 kg mass moving in the
Nat2105 [25]

Answer:

Explanation:

mdeemmkdkwdmwdmw

4 0
3 years ago
Make a table table that shows the causes and effects of local winds and monsoons
kiruha [24]
I'd answer that but I can't text graphs and tables...
4 0
3 years ago
A garage door opener has a power rating of 350 watts.
Lelechka [254]
We Know, Power = Energy/Time
Substitute the known value. which is P = 350 watt, & T = 30 sec
350 = E/30 
E = 350 * 30
E = 1050 Joules.

So, your answer is 1050 Joules.
Hope this helps! Keep studying! 
6 0
3 years ago
Read 2 more answers
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