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3 years ago

The first four steps in the formation of the sun are listed below.

2 answers:

7 0

<span>The next logical step in the formation of the sun would be choice B. Nuclear fusion of gases takes place in the center of the nebula. </span>

Airida [17]3 years ago

6 0

(B)Option b is your answer..........

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Please help with any of them!!!!!!

Law Incorporation [45]

Its to blury for me to see it

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3 years ago

You and a friend both leave the same restaurant to drive home. You are heading directly west at 30 miles per hour and he or she

steposvetlana [31]

Answer:

$\frac{dAB}{dt} = = 50 miles/ hour$

Explanation:

let A represent me and B represent my friend

A speed $\frac{dOA}{dt}$ =30 m/hr toward west

B speed $\frac{dOB}{dt}$ =40 m/hr toward south

after 1/2 hr

total distance cover by A =1/2 * 30 miles = 15 miles

total distance by B = 1/2*40 miles = 20 miles

now from figure

$AB = \sqrt{(15^2+20^2)} = 25 miles$

$AB^2 =OA^2+ OB^2$

Differentiate above equation

$2AB\frac{dAB}{dt} = 2OA\frac{dOA}{dt} +2OB\frac{dOB}{dt}$

$25*\frac{dAB}{dt} = 15*30 +20*40$

$\frac{dAB}{dt} = \frac{1250}{25}$

$\frac{dAB}{dt} = = 50 miles/ hour$

5 0

3 years ago

Read 2 more answers

A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri

vodka [1.7K]

We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2

5 0

2 years ago

Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20

Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

$t=102.8 minutes$

Explanation:

Given: $m=500kg$ , $r=1.0m$ , $w=200\pi rad/s$

a). The rotational kinetic energy

$K_R=\frac{1}{2}*I*w^2$

$I=\frac{1}{2}*m*r^2$

$I=\frac{1}{2}*500kg*(1.0m)^2$

$I=250 kg*m^2$

$K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2$

$K_R=49.348x10^6J$

b). The power average 0.8kW un range time can be find

$P=\frac{K_R}{t'}$

Solve to t'

$t=\frac{K_R}{P}$

$t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s$

$t=6168.5s\frac{1minute}{60s}=102.8 minutes$

3 0

3 years ago

When the heat source is removed from a fluid, convection currents in the fluid will eventually ____________then stop.

Blizzard [7]

When the heat source is removed from a fluid, convection currents in the fluid will eventually distribute heat uniformly throughout the fluid. When all of the fluid is at the same temperature, convection currents will stop.

5 0

3 years ago