Questioning is the beginning of the scientific inquiry process. True or False

The volume of water in the Pacific Ocean is about 7.00 × 108 km3. The density of seawater is about 1030 kg/m3. For the sake of t

oee [108]

The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.

By definition the gravitational potential energy is given by,

$PE=\frac{GMm}{r}$

Where,

m = Mass of Moon

G = Gravitational Universal Constant

M = Mass of Ocean

r = Radius

First we calculate the mass through the ratio given by density.

$m = \rho V$

$m = (1030Kg/m^3)(7*10^8m^3)$

$m = 7.210*10^{11}Kg$

PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon

Now we define the radius at the most distant point

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

Then the potential energy at this point would be,

$PE_1 = \frac{GMm}{r_1}$

$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

$PE_1 = 9.05*10^{15}J$

PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.

At the nearest point we perform the same as the previous process, we calculate the radius

$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

The we calculate the Potential gravitational energy,

$PE_2 = \frac{GMm}{r_2}$

$PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}$

$PE_2 = 9.361*10^{15}J$

7 0

3 years ago

A suitcase is pulled 28 feet along a flat sidewalk with a constant upward force of 80 lb at an angle of 23degrees with the horiz

Elenna [48]

Answer:

2795.3 J

Explanation:

distance, d = 28 feet = 8.53 m

Force, F = 80 lb = 356 N

Angle, θ = 23°

Work = F x d x cos θ

W = 356 x 8.53 x cos 23

W = 2795.3 J

Thus, the work done is 2795.3 J.

5 0

4 years ago

Three perfectly polarizing sheets are spaced 2?cm {\rm cm} apart and in parallel planes. The transmission axis of the second she

Agata [3.3K]

Answer:

I₃/Io % = 0.8.59

Explanation:

A polarizer is a complaint sheet for light in the polarization direction and blocks the perpendicular one. When we use two polarizers the transmission between them is described by Malus's law

I = I₀ cos² θ

Let's apply the previous exposures in our case, the light is indicatively not polarized, so the first polarized lets half of the light pass

I₁ = ½ I₀

The light transmitted by the second polarizer

I₂ = I₁ cos² θ

I₂ = (½ I₀) cos2 28

The transmission by the polarizing third is

I₃ = I₂ cos² θ₃

The angle of the third polarizer with respect to the second is

θ₃ = 90-28

θ₃ = 62º

I₃ = (½ I₀ cos² 28 cos² 62)

Let's calculate

I₃ = Io ½ 0.7796 0.2204

I₃ = Io 0.0859

I₃/Io= 0.0859 100

I₃/Io % = 0.8.59

4 0

4 years ago

A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the

MArishka [77]

Answer:

The the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is

$I_c = \frac{I}{I_o} =8.48 *10^{-8}$

Explanation:

From the question we are told that

The width of the slit is $D = 0.3 \ mm = 0.3 *10^{-3} \ m$

The wavelength is $\lambda = 254 \ nm = 254 *10^{-9} \ m$

The angle is $\theta = 11^o$

The intensity of at $11^o$ to the axis in terms of the intensity of the central maximum. is mathematically represented as

$I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2$

Where $\beta$ is mathematically represented as

$\beta = \frac{D sin (\theta ) * \pi}{\lambda }$

substituting values

$\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }$

$\beta = 708.1 \ rad$

So

$I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2$

$I_c = \frac{I}{I_o} =8.48 *10^{-8}$

6 0

4 years ago

Two hypothetical discoveries in Part A deal with moons that, like Earth's moon, are relatively large compared to their planets.

arlik [135]

Answer:

Unusually large moons form in giant impacts, which are relatively rare events

Explanation:

Solution:

- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.

- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible. The transition from an undifferentiated cloud to a star system complete with planets and moons takes about 100 million years.

8 0

3 years ago