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Aloiza [94]
3 years ago
13

Which of the following planets may have a habitable moon? PLZ ANSWER

Physics
2 answers:
Black_prince [1.1K]3 years ago
8 0

Answer:

Jupiter

Saturn

Explanation:

Yuki888 [10]3 years ago
3 0
The moon calisto which orbits the planet Jupiter
Answer-Jupiter
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MRU(movimiento rectilineo uniforme) un movil viaja en linea recta con una velocidad media de 12metros /segundos durante 9segundo
liubo4ka [24]

Answer:

a) 141.6m

b) 8.4m/s

Explanation:

a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:

x_1=v_1t_1=(12m/s)(9s)=108m\\\\x_2=v_2t_2=(4.8m/s)(7s)=33.6m\\\\x_T=x_1+x_2=108m+33.6m=141.6m

hence, the total distance is 141.6m

b) the mean velocity of the total trajectory is given by:

v_m=\frac{v_1+v_2}{2}=\frac{12m/s+4.8m/s}{2}=8.4\frac{m}{s}

hence, the mean velocity is 8.4 m/s

8 0
3 years ago
lcompute the increase in length of 500 m of copper wire when it's temperature changes from 12 degrees celsius to 32 degrees Cels
marshall27 [118]

Explanation:

step by step explanation

7 0
3 years ago
How many significant figures does 0.00340 have?
True [87]
Greetings!

"<span>How many significant figures does 0.00340 have?"...

A significant figure is:
-A non zero number
-A zero in between non zero numbers
-Trailing zeros to the right of the decimal point

Therefore, this number has three significant figures:
3,4,0

Hope this helps.
-Benjamin</span>
4 0
3 years ago
Read 2 more answers
Differences between Light year and Cosmic year in two points
rusak2 [61]

Answer:

A cosmic year is 365.25 days, some times called a side real year and is just the time it takes for us to go round the sun once.

A light year is the distance light travels in a year. Now light travels at about 186,262 miles a Second! Which is not slow by any ones book.

An experiment was conducted just after Christmas a few years ago. Two girls were selected from the audience and went into two phone boxes a few feet apart. They could only hear each other via the phones. The phone call went to a ground station about 200 miles away, then up to a geostationary coms satellite, back to a ground station 1/3 of the way around the world, then repeated, with a third satellite before being sent from another ground station back to London and the other phone box. We the audience could hear both sides of the conversation from both boxes. And could hear the delay between sending and receiving. So even at the speed of light, there was about 1.5 seconds of delay. So because distances in space are so vast that saying a star is x millions of miles away causes problems, you run out of zero’s! So our nearest other star is about 4.5 light years away. Our sun (our nearest start) is about 8 light minuets away. Varies slightly as our orbit is not 100% cirular.

I HOPE THIS IS HELPFUL.

3 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
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