0.002 written in scientific notation

Do Earth’s oceans gain or lose water, considering evaporation and precipitation together? How much?

lianna [129]

Much of the precipitation in large bodies of water occurs at the surface. The ocean loses about 37000 km cubed considering evaporation and precipitation.

7 0

3 years ago

A scientist compares two samples of colorless gas present at the beginning and end of a lab process. She wants to determine whet

kompoz [17]

Joseph Priestley was born in Yorkshire, the eldest son of a maker of wool cloth. His mother died after bearing six children in six years. Young Joseph was sent to live with his aunt, Sarah Priestley Keighley, until the age of 19. She often entertained Presbyterian clergy at her home, and Joseph gradually came to prefer their doctrines to the grimmer Calvinism of his father. Before long, he was encouraged to study for the ministry. And study, as it turned out, was something Joseph Priestley did very well.

Aside from what he learned in the local schools, he taught himself Latin, Greek, French, Italian, German and a smattering of Middle Eastern languages, along with mathematics and philosophy. This preparation would have been ideal for study at Oxford or Cambridge, but as a Dissenter (someone who was not a member of the Church of England) Priestley was barred from England's great universities. So he enrolled at Daventry Academy, a celebrated school for Dissenters, and was exempted from a year of classes because of his achievements.

After graduation, he supported himself, as he would for the rest of his life, by teaching, tutoring and preaching. His first full-time teaching position was at the Dissenting Academy in Warrington. (Although obviously brilliant, original, outspoken and, by one report, of "a gay and airy disposition," Priestley had an unpleasant voice and a sort of stammer. That he made a living through lectures and sermons is further evidence of his extraordinary nature.)

https://www.acs.org/content/acs/en/education/whatischemistry/landmarks/josephpriestleyoxygen.html

8 0

3 years ago

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1 kg ball rolls off a 33 m high cliff, and lands 23 m from the base of the cliff. Express the displacement and the gravitational

enyata [817]

Answer:

d = <23, 33, 0> m , F_W = <0, -9.8, 0> , W = -323.4 J

Explanation:

We can solve this exercise using projectile launch ratios, for the x-axis the displacement is

x = vox t

Y Axis

y = $v_{oy}$ t - ½ g t²

It's displacement is

d = x i ^ + y j ^ + z k ^

Substituting

d = (23 i ^ + 33 j ^ + 0) m

Using your notation

d = <23, 33, 0> m

The force of gravity is the weight of the body

W = m g

W = 1 9.8 = 9.8 N

In vector notation, in general the upward direction is positive

W = (0 i ^ - 9.8 j ^ + 0K ^) N

W = <0, -9.8, 0>

Work is defined

W = F. dy

W = F dy cos θ

In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º

Cos 180 = -1

W = -F y

W = - 9.8 (33-0)

W = -323.4 J

6 0

3 years ago

Anino acids are important to the body because they build

tino4ka555 [31]

Answer:

Amino acids and proteins are the building blocks of life. When proteins are digested or broken down, amino acids are left. The human body uses amino acids to make proteins to help the body: Break down food.

6 0

3 years ago

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A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz

gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle? Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa

Tt is stagnation temperature = ?

T is the static temperature = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa = ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 = Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0

3 years ago