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Elodia [21]
3 years ago
10

What is the name for the area of dust and debris which orbited the young sun and eventually became planets and moons?

Physics
2 answers:
Talja [164]3 years ago
6 0
The answer to this question would be C. <span>protoplanatary disc</span>

(According to dictionary.com)- A rotating disk of dust and gas that surrounds the core of a developing solar system. It may eventually develop into orbiting celestial bodies such as planets and asteroids.
alexandr402 [8]3 years ago
5 0
I believe the answer is C. protoplanetary disc. Let me know if this helps
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5. A helicopter scoops a load of water out of a lake, to dump on a forest fire. The helicopter exerts 6480
Strike441 [17]

Answer:

h = 250 m

Explanation:

Given that,

The helicopter exerts 6480  N of force on the water while rising high enough to fly over a mountain.

The helicopter does 1.62\times 10^6\ J of work on the water lifting it.

We need to find how high does it lift the water. Work done is given by :

W=Fh\\\\h=\dfrac{W}{F}\\\\h=\dfrac{1.62\times 10^6}{6480}\\\\h=250\ m

So, it can lift the water to a height of 250 m.

6 0
3 years ago
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
4 years ago
Explain why squeezing a plastic mustard bottle forces mustard out to top.
Oliga [24]

Answer:

because when squeezing you are increasing pressure within the bottle and there is less pressure on the outside

5 0
3 years ago
Read 2 more answers
What force is needed to accelerate a 0.5kg football at a rate of 40m/s
GREYUIT [131]
Force=mass x acceleration
f= 0.5 x40
f=20N
6 0
3 years ago
In which of the following examples does the object have both kinetic and potential energy? Select all that apply.
notsponge [240]
I believe the answer is H for when you bounce it, it has stress when it hits the floor and then goes up giving it kinetic
6 0
3 years ago
Read 2 more answers
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