3 years ago

How does making models help scientists observe?

2 answers:

8 0

By creating models scientists can have a little knowledge how specific element might look, and by the way will help them to be more creative, imaginative and reasonable.

Oksanka [162]3 years ago

7 0

Well it helps them haveing a model so they can discover new observations on life and everything around them. hope it helped

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You weigh 580 N on Earth. If you were to go to Mars, where its gravitational pull is 3 . 7 11 m /s 2 , what would you weigh? (Hi

andrew-mc [135]

Answer:

59.18 kg

Explanation:

use f=ma

f= 580 N

a = 9.8 m/s 2

weigh(m) doesn't change only force(F) changes

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2 years ago

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An astronaut weighs 104 newtons on the moon, where the strength of gravity is 1.6N/kg. What is her mass?

Leona [35]

Her mass is 65 kilograms

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3 years ago

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I need help with number 6

Otrada [13]

I like your handwriting boy

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3 years ago

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

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3 years ago

The diagram below shows the velocity vectors for two cars that are moving relative to each other.

scoundrel [369]

Answer:

The answer is " $5 \ \frac{m}{s} \ west$ "

Explanation:

$\to \vec{V_1} = (25 \frac{m}{s}) (\hat{-i})\\\\\to \vec{V_2} = (20 \frac{m}{s}) (\hat{-i})\\\\$

velocity of car | respect to car :

$\to \vec{V_{12}} = \vec{V_1} - \vec{V_2}\\\\$

$=\vec{-25} \hat{i}+ \vec{20} \hat{i}\\\\= 5 \ \frac{m}{s} \ west$

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3 years ago