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4 years ago

13

Suppose that the clay balls model the growth of a planetesimal at various stages during its accretion. Choose the planetesimal t

hat is most likely
to pull in debris particle A.

1 answer:

motikmotik4 years ago

8 0

Answer:2

Explanation:

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b) A satellite is in a circular orbit around the Earth at an altitude of 1600 km above the Earth's surface. Determine the orbita

asambeis [7]

Explanation:

The orbiting period of a satellite at a height h from earth' surface is

T=2πr32gR2

where r=R+h.

Then, T=2π(R+h)R(R+hg)−−−−−−−−√

Here, R=6400km,h=1600km=R/4

T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√

Putting the given values,

T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=7092s=1.97h

Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is 24h. Let us find the height h of such a satellite above the earth's suface in terms of the earth,'s radius.

Let it be nR.Then

T=2π(R+nR)R(R+nRg)−−−−−−−−−−√

=2π(Rg)−−−−−√(1+n)32

=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32

(5075s)(1+n)32=(1.41h)(1+n)32

For T=24h, we have (24h)=(1.41h)(1+n)32

or (1+n)32=241.41=17

or 1+n(17)23=6.61

or n=5.61

The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.

3 0

3 years ago

Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc

Lapatulllka [165]

Answer:

a) d = 30.79 m , b) θ = -22.4° , θ = 22.4 South of East

Explanation:

The easiest way to solve problems with vectors is to use their components, for this the East-West direction coincides with the x-axis and the North-South direction coincides with the y-axis

Let's use the index for / Ricardo and the index for Jane, let's break down the displacements

Richard

X axis

x₁ = 26.0 sin (60)

x₁ = -22.52 m

Y Axis

y₁ = 26.0 cos 60

y₁ = 13 m / s

Jane

X axis

x₂ = 16.0 cos (180 +30)

x₂ = -13.85 m

Y Axis

y₂ = 16.0 sin (180 + 30)

y₂ = - 8.0 m

Now we can use Pythagoras' theorem to find the distance between them

d = √ [(x₂ -x₁)² + (y₂ -y₁)²]

d = √ [(-13.85 + 22.52)² + (-8 -13)²]

d = 30.79 m

Let's use trigonometry to enter the address

tan θ = Δy / Δx

θ = tan⁻¹ Δy / Δx

θ = tan⁻¹ (-13.85 + 22.52) / (-8 - 13)

θ = tan⁻¹ (-8.67 / 21)

θ = -22.4°

The negative sign indicates that the angle is measured from the axis clockwise.

In the form of cardinal s point is

θ = 22.4 South of East

4 0

4 years ago

A car with a velocity of 22 m/s is accelerated at a rate of 1.6m/s2 for 6.8s. determine the final velocity

Ivahew [28]

A car with a velocity of 22 m/s is accelerated at a rate of 1.6 $m/s^2$ for 6.8s has the final velocity t be 32.88 m/s.

The acceleration means the amount of velocity changing per unit time.

The given data:

initial velocity, u = 22 m/s

time, t = 6.8 s

acceleration, a = 1.6 $m/s^2$

We will be using the equation of motion:

v = u + at

$\therefore v=22+1.(6.8)$

$\Rightarrow v=22+10.88$

$\Rightarrow v=32.88 \ m/s$

The final velocity become 32.88 m/s.

To learn more about Attention here:

https://brainly.in/question/10557838

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3 0

2 years ago

Which of the following best describes gravitational potential energy,

s2008m [1.1K]

The energy stored in an object when you lift it,

8 0

3 years ago

A neutral wire without current flowing creates A.) an electric field. B.) a magnetic field. C.) both an electric field and magne

Wewaii [24]

Believe the answer is B

6 0

3 years ago