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Greeley [361]
3 years ago
12

What are two possible paths of a spacecraft that just launched from Earth but that is not traveling at orbital velocity?

Biology
2 answers:
beks73 [17]3 years ago
8 0
A spacecraft that is not traveling at an orbital velocity will:

a. get attracted into a different kind of gravity or pole
b. will become off-set and be lost within space

This is the main reason why spacecrafts are well computed mathematically and engineering wise in order for it to travel and perform its roles in space.
user100 [1]3 years ago
8 0

Answer:

Orbital velocity can be described as the velocity at which the spacecraft would orbit the Earth at a fixed distance.

v=\sqrt{\frac{GM}{R}}

Where,

G is the Gravitational constant

M is the mass of the Earth

R is the distance from the Earth's center

If the spacecraft is not launched at orbital velocity then it would not orbit the Earth. The spacecraft could fall back to the Earth or could go into space.

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❥SCIENCE SUBJECT
Fiesta28 [93]

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Answer:

  • time: 1.122 seconds
  • range: 10.693 m
  • maximum height: 1.543 m

Explanation:

<u>Given</u>:

  runner is launched at 30° angle to horizontal at 11 m/s

  acceleration due to gravity is g = -9.8 m/s²

<u>Find</u>:

  runner's hang time

  runner's distance to the landing point

  runner's maximum height

<u>Solution</u>:

The (horizontal, vertical) speed components will be ...

  (11 m/s)(cos(30°), sin(30°)) = (5.5√3 m/s, 5.5 m/s)

The time of flight can be found from the height formula:

  h(t) = 1/2gt² +vt . . . . . . where v is the vertical speed at launch

The time we're concerned with is the time when h(t)=0 and t>0.

  0 = -4.9t^2 +5.5√3t = t(-4.9t +5.5√3)

The second factor is zero when ...

  t = (5.5√3)/4.9 ≈ 1.122 . . . seconds hang time

__

The distance to the landing point will be the product of horizontal speed and hang time:

  d = (5.5 m/s)(5.5√3/4.9 s) ≈ 10.693 m . . . . distance to landing

__

The maximum height can be found from the formula (based on conversion of kinetic energy to potential energy) ...

  h = v²/|2g| = (5.5 m/s)²/(2(9.8 m/s²)) ≈ 1.543 m . . . . maximum height

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In sickle-cell disease, as a result of a single amino acid change, the hemoglobin tetramers associate with each other and assemb
Phoenix [80]

Answer:

One of the common genetic disorders is sickle cell anemia, in which 2 recessive alleles must meet to allow for destruction and alteration in the morphology of red blood cells. This usually leads to loss of proper binding of oxygen to hemoglobin and curved, sickle-shaped erythrocytes. The mutation causing this disease occurs in the 6th codon of the HBB gene encoding the hemoglobin subunit β (β-globin), a protein, serving as an integral part of the adult hemoglobin A (HbA), which is a heterotetramer of 2 α chains and 2 β chains that is responsible for binding to the oxygen in the blood. This mutation changes a charged glutamic acid to a hydrophobic valine residue and disrupts the tertiary structure and stability of the hemoglobin molecule. Since in the field of protein intrinsic disorder, charged and polar residues are typically considered as disorder promoting, in opposite to the order-promoting non-polar hydrophobic residues, in this study we attempted to answer a question if intrinsic disorder might have a role in the pathogenesis of sickle cell anemia. To this end, several disorder predictors were utilized to evaluate the presence of intrinsically disordered regions in all subunits of human hemoglobin: α, β, δ, ε, ζ, γ1, and γ2. Then, structural analysis was completed by using the SWISS-MODEL Repository to visualize the outputs of the disorder predictors. Finally, Uniprot STRING and D2P2 were used to determine biochemical interactome and protein partners for each hemoglobin subunit along with analyzing their posttranslational modifications. All these properties were used to determine any differences between the 6 different types of subunits of hemoglobin and to correlate the mutation leading to sickle cell anemia with intrinsic disorder propensity.

Explanation:

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