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3 years ago

How would the absence of gravity affect the formation of planets?

Physics

2 answers:

Doss [256]3 years ago

5 0

Answer : The rocks and debris would never accrete into a planet.

Explanation : Gravity is an attraction force and dependents on the mass of the objects but in the absence of the force the rocks and debris would not come together to form a clump.

So, we can say that in the absence of gravity, the rock and debris would never accrete into a planet.

Maru [420]3 years ago

3 0

The correct answer is B. The rocks and debris would never accrete into a planet.

Explanation:

Gravity refers to a force of attraction between objects due to their masses. This force is described through the formula F= G M m/ r2 in which G is equal to the gravitational constant, M and m describe the mass of objects, and r refers to the distance.

Moreover, gravity was essential for the formation of planets, stars and other elements in the universe, because this caused gas, dust, rocks, and other particles or elements to aggregate or accrete into planets and other celestial bodies. Moreover, gravity is responsible for phenomena such as planets orbiting. The previous idea implies that without gravity the formation of planets would not have possible and therefore "the rocks and debris would never accrete into a planet".

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Light intensity is?

Fittoniya [83]

The correct answer is:
<span>The rate at which a waves energy flows through a given unit of area

In fact, light intensity is defined as the light power per unit of area:
</span> $I= \frac{P}{A}$
<span>but the power is the energy carried by the light per unit of time:
</span> $P= \frac{E}{t}$
<span>this means that the intensity can be rewritten as
</span> $I= \frac{E}{tA}$ <span>
So, it's basically the rate of energy (per unit of time) through a given surface.</span>

5 0

3 years ago

Read 2 more answers

A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction

lara31 [8.8K]

<h3>Answer</h3><h3>7 Ns</h3><h3>Explanation</h3>

Given in the question,

mass of foul ball = 0.140 kg

initial speed with which ball was hit with the bat = 30 m/s

final speed = 40 m/s

According to the scenario the whole scene is making a right angle triangle

So, to the solve the question we will use pythagorus theorem

<h3>Hypotenuse² = base² + height²</h3>

Here,

Hypotenuse= Magnitude of impulse

Base = 1st change of momentum

height = 2nd change of momentum

1st impulse (1st change of momentum)

p = m(1)v(1) = (0.14 kg)(40.0 m/s) = 5.6 kg m / s = 5.6 N s

2nd impulse (2nd change of momentum)

p = m(2)v(2) = (0.14 kg)(30.0 m/s) = 4.2 kg m / s = 4.2 N s

Magnitude of impulse (hypotenuse of triangle)

impulse² = (5.6)² + (4.2)²

impulse² = 31.36 + 17.64

impulse² = 49

impulse² = √49

impulse = 7.0 N s

7 0

3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

ELEN [110]

With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$