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3 years ago

Compare and contrast scientific inquiry "skill and process".

Physics

1 answer:

Dmitrij [34]3 years ago

3 0

Answer:

The inquiry process takes advantage of the natural human desire to make sense of the world... This attitude of curiosity permeates the inquiry process and is the fuel that allows it to continue. Process skills are not used for their own sake.

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Examples of drawing packages

Marat540 [252]

Answer:

The answer are given above in attachment.

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3 years ago

Afootball of mass 550g is at rest on the ground the football is kicked with a force of 108 newton the footballers boot is in con

d1i1m1o1n [39]

If "0.3 minute" is correct, then it's 9,543,272 Joules.

If it's supposed to say "0.3 SECOND", then the KE is 2,651 Joules.

6 0

3 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

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The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago

A compound has a melting point of 812°C. what type of bonding is in the compound?

AURORKA [14]

Answer:

Ionic compound.

Explanation:

812° C is a very high melting point. Such high melting points are generally ionic compound. Ionic compounds are have very strong bond between the elements ( electrostatic bond). In order to break this bond, large amount of heat energies are needed. So, they have high melting point. Also, Ionic compound are very good conductors of electricity.

7 0

3 years ago

Natalie lifts a 15-kg rock from the ground onto a 1.5 meter high wall. what is the amount of potential energy she has given the

Zina [86]

The amount of gravitational potential energy acquired by the rock is equal to:
$\Delta U = mg \Delta h$
where
m is the mass of the rock
g is the gravitational acceleration
$\Delta h$ is the increase in height of the rock

Substituting the data of the problem, we find
$\Delta U=(15 kg)(9.81 m/s^2)(1.5 m)=220.7 J$
So, Natalie gave 220.7 J of energy to the rock.

4 0

3 years ago