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tatuchka [14]
3 years ago
12

How often do you rely on media for scientific information to make decisions in your life? (No weather answer please)

Physics
2 answers:
vodka [1.7K]3 years ago
6 0
The majority of the time


Nina [5.8K]3 years ago
4 0
Well, taking out the weather option ... I hardly ever do.
If I'm looking at a specific technical problem, then a scan on a source I'm not allowed to cite here, it seems, or some such may give me a hint as to whether it's going to be any help to me. If it is, i continue. If it's isn't, then i'll go away and decide whether I should pursue the problem or not (ie am I wasting my energy here ?)
I suspect, though, that I'm not representative of users.
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A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare
zvonat [6]

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

F = \frac{\Delta P}{\Delta t}

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

F_{12} = F_{21}

so we know that

\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}

so we have

\Delta P_{12} = \Delta P_{21}

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

6 0
3 years ago
The brakes on a 15,680 N car exert a stopping force of 640 N.
Murrr4er [49]

Answer:

1568 Kg is the answer

Explanation:

6 0
3 years ago
What gravitational force does the earth exert on a person
lara [203]

The  gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

<h3>What is the gravitational force of the earth on the person?</h3>

The gravitational force exerted by the earth on a person standing on the earth's surface is given below as follows:

  • F = \frac{Gm^{1}m^{2}}{r^{2}}

where

G = 6.67 * 10⁻¹¹

m¹ = 62 kg

m² = 5.97 * 10²⁷ kg

r = 6.4 * 10⁶ m

F = \frac{5.97*10^{24}*62*6.67*10^{-11}}{(6.4*10^{6}){2}} = 602.74\:N

Therefore, the gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

Learn more about gravitational force at: brainly.com/question/940770

#SPJ1

7 0
2 years ago
Describe how the ocean influences climate by storing heat and water? Give one example to justify your answer.
just olya [345]
The sun light that received by the water in the ocean will increase the average temperature of the water and make it warmer.
With the help of wind and current, the warm water will spread out to another region, and increasing the average temperature in that region and affecting its overall climate.
Example The temperature rise in Valdivia<span>, Chile and in Beijing, China after receiving warm water from arctic.</span>
7 0
3 years ago
Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr
vivado [14]

This question can be solved from the Kepler's law of planetary motion.

As per this law the square of time period of a planet  is proportional to the cube of semi major axis.

Mathematically it can be written as   T^{2} \alpha R^{3}

                                                          ⇒T^{2} = KR^{3}

Here K is the proportionality constant.

If T_{1} andT_{2} are the orbital periods of the planets and

R_{1} and R_{2} are the distance of the planets from the sun, then Kepler's law can be written as-

          \frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{2} }

      ⇒ R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }

  Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.

Let the distance from sun and orbital period of Saturn is denoted as R_{1} and T_{1} respectively.

Let the distance  from sun and orbital period of earth is denoted as R_{2} and T_{2} respectively.

we are given thatT_{1} =29.46 years

we know that R_{2} = 1 AU and T_{2} = 1 year.

1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.

Hence distance of Saturn from sun  is calculated as -

From Kepler's law as mentioned above-

                                    R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }

                                             =[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU

                                    =867.8916 AU^{3}

                                        ⇒R_{1} =\sqrt[3]{867.8916}

                                           =9.5386 AU [ans]

5 0
3 years ago
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