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Schach [20]
3 years ago
15

What are the three reasons why nebula contribute more to Stellar formation than any other region in the universe?

Physics
2 answers:
monitta3 years ago
7 0
I think it is a, b, and e. Hope it helps! :)
kati45 [8]3 years ago
5 0

Answer:

A D E

Explanation:

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In the formula Work=Force X Distance, what happens to the amount of work done if either Force or Distance is increased?
AfilCa [17]

Explanation:

If force or distance is increased, then amount of workdone will also increase.

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A motorist wishes to travel 40 kilometers at an average speed of 40 km/h. During the first 20 kilometers, an average speed of 40
allochka39001 [22]

poste en français s’il vous plaît

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A stream begins at an elevation of 1000 feet and flows a distance of 200 miles to the ocean. what is the average gradient?
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3 years ago
A wagon is rolling forward on level ground. friction is negligible. the person sitting in the wagon is holding a rock. the total
vladimir2022 [97]

use the equation mv+mv=mv+mv

95.5(.51)= 95.2(X)+ 17.5(.290)

Now solve for X and that should be your answer

OR

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7 0
3 years ago
In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient
morpeh [17]

Answer:

v(t) = 21.3t

v(t) = 5.3t

v(t) = 48 -48 e ^{ \frac{t}{9}}

Explanation:

When no sliding friction and no air resistance occurs:

m\frac{dv}{dt} = mgsin \theta

where;

\frac{dv}{dt} = gsin \theta , 0 < \theta <  \frac{ \pi}{2}

Taking m = 3 ; the differential equation is:

3 \frac{dv}{dt}= 128*\frac{1}{2}

3 \frac{dv}{dt}= 64

\frac{dv}{dt}= 21.3

By Integration;

v(t) = 21.3 t + C

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta

Taking m =3 ; the differential equation is;

3 \frac{dv}{dt}=128*\frac{1}{2} -\frac{\sqrt{3} }{4}*128*\frac{\sqrt{3} }{4}

\frac{dv}{dt}= 5.3

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta - kv

The differential equation is :

= 3 \frac{dv}{dt}=128*\frac{1}{2} - \frac{ \sqrt{ 3}}{4}*128 *\frac{ \sqrt{ 3}}{2}-\frac{1}{3}v

= 3 \frac{dv}{dt}=16 -\frac{1}{3}v

By integration

v(t) = 48 + Ce ^{\frac{t}{9}

Since; V(0) = 0 ; Then C = -48

v(t) = 48 -48 e ^{ \frac{t}{9}}

7 0
3 years ago
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