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3 years ago

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A cosmic catastrophic event occurred that caused the tilt of the Earth's axis relative to its plane of orbit to increase from 23

.5 degrees to 90 degrees. The most obvious effect of this change would be

1 answer:

Gnom [1K]3 years ago

7 0

Answer: The elimination of seasonal variations

Explanation:

Since the cosmic catastrophic event which occurred led to the tilt of the Earth's axis relative to the plane of orbit to increase from 23.5° to 90°, the most obvious effect of this change would be the elimination of seasonal variations.

It should be noted that seasonal variation refers to the variation in a time series that's within a year which is repeated. The cause of seasonal variation can include rainfall, temperature, etc.

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Why is it expensive to bring electricity into urban areas?

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Because of the pole and the generator you would have to biuld

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3 years ago

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A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

3 years ago

One gallon of paint (volume = 3.79 10-3 m3) covers an area of 17.8 m2. What is the thickness of the fresh paint on the wall?

AURORKA [14]

Answer:

0.2129 mm

Explanation:

We have given volume of the paint = 1 gallon $=3.79\times 10^{-3}m^3$

Area that covers the paint $=17.8 m^2$

We have to find the thickness of the fresh paint

So $thickness=\frac{volume}{area}=\frac{3.79\times 10^{-3}}{17.8}=0.2129\times 10^{-3}m=0.2129mm$

So the thickness of fresh paint on the wall is 0.2129 mm

6 0

3 years ago

If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.

grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution. By definition is defined as:

$\theta = 1.22\frac{\lambda}{d}$

Where,

$\lambda$ = Wavelength

d = Width of the slit

$\theta$ = Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

$r = l\theta$

Relacing $\theta$

$r = l(\frac{1.22\lambda}{d})$

$r = 1.22\frac{l\lambda}{d}$

Replacing with our values we have that,

$r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})$

$r = 3680.91m$

Therefore the diameter of the beam on the moon is

$d = 2r$

$d = 2 * (3690.91)$

$d = 7361.8285m$

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0

3 years ago

A block of gelatin is 120 mm by 120 mm by 40 mm when unstressed.

ycow [4]

Answer:

σ = 3.402 KPa , γ = 0.25 , G = 13.608 KPa

Explanation:

Given:-

- The dimension of gelatin block = ( 120 x 120 x 40 ) mm

- The applied force, F = 49 N

- The displacement of upper surface, x = 10 mm

Find:-

Find the shearing stress, shearing strain and shear modulus.

Solution:-

- The shear stress is the internal pressure created in an object opposing the applied action ( Force, moment, bending, or torque ).

- A force of F = 49 N was applied parallel to the top surface of the gelatin block.

- The shear effect results in a stress in the gelatin block.

- The formulation of stress ( σ ) is given below:

σ = F / A

Where,

A : The surface area of the object that experiences the shear force.

- The top surface have the following dimensions:

A = ( 0.120 )*( 0.120 ) = 0.0144 m^2

Therefore,

σ = 49 / 0.0144

σ = 3.402 KPa

- The shear strain ( γ ) is the measurement of change in dimension per unit depth of the block.

- The top surface undergoes a displacement of ( x ). The height of the top surface of the gelatin block is L = 40 mm.

Hence,

γ = x / L

γ = 10 / 40

γ = 0.25

- The shear modulus or the modulus of rigidity ( G ) is a material intrinsic property that signifies the amount of resistive stress to any cause of deformation.

- It is mathematically expressed as a ratio of shear stress ( σ ) and shear strain ( γ ):

G = σ / γ

G = 3.402 / 0.25

G = 13.608 KPa

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3 years ago