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2 years ago

15

The solar mass of the Sun is 1. The orbital period of Jupiter is 11. 9 Earth years. What is the distance between Jupiter and the

Sun? 5. 2 AU 41 AU 52 AU 410 AU.

1 answer:

cestrela7 [59]2 years ago

6 0

The distance between Jupiter and the sun is 5.2 AU.

According to Kepler's third law, the square of the period of revolution of planets is proportional to the cube of their mean distances from the sun. From this; T^2 = r^3.

Now, we are told that the orbital period (T) is 11. 9 Earth years. We have to make the distance the subject of the formula.

r =T^2/3

r = (11.9)^2/3

r = 5.2 AU

Learn more: brainly.com/question/15207516

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Answer:

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3 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago

Suppose you exert a 25-N force to lift a ball 0.4 m in 2 s. How much work is done?

Sergio [31]

work is force x distance = 25 x 0.4

= 2.5x4 = 10joules

pwer would be 10j/2s watts .... 5 watts

3 0

3 years ago

parallel-plate air capacitor is made from two plates 0.070 m square, spaced 6.3 mm apart. What must the potential difference bet

Rom4ik [11]

Answer:

V = 576 V

Explanation:

Given:

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Find:

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Solution:

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u = 0.5*ε*E^2

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Hence,

u = 0.5*ε*(V/d)^2

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Plug in values:

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4 0

3 years ago

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Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

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$\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}$

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Using Newton's second law, it is established that F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object $\left(m_{1}+m_{2}\right)$

Now consider both $\mathrm{m}_{1} \text { and } \mathrm{m}_{2}$ as a system, so net force acting on the system is $\text { Force }=\left(m_{1}+m_{2}\right) a$

Substitute the given values in the above formula,

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Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

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3 years ago