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disa [49]
2 years ago
10

Gravity pulled materials together forming the planets in our solar system. The inner planets consist of dense material, such as

iron, which was pulled to the center of the planets. Materials such as ice, liquid and gas formed the outer planets. Scientist identify this process as:.
Physics
1 answer:
Ede4ka [16]2 years ago
7 0

Scientist identify the process in which gravity pulled materials together forming the planets in our solar system as accretion.

Accretion is the formation model of planets in our solar system in which gravity pulls materials together forming the planets.

This formation model accounts for the formation of the inner planets made of of dense material such as iron. These planets include Mercury, venus, earth etc.

As the gravity pulls in the more dense material, the less dense material such as ice, liquid and gas colect together to form the outer planets, which include Jupiter, Uranus, etc.

So, Scientist identify the process in which gravity pulled materials together forming the planets in our solar system as accretion.

Learn more about accretion here:

brainly.com/question/17392929

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A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its
m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0
3 years ago
What is the wavelength of violet light that has a frequency of 7.5x10^14
Crank
Λ= V/f 
<span>but change it to represent the speed of light, c </span>
<span>λ= c/f </span>
<span>c = 3.00 x 10^8 m/s </span>
<span>Plug in your given info and solve for λ(wavelength) </span>
<span>λ= 3.00 x 10^8 m/s / 7.5 x 10^14 Hz
(3.00 x 10^8) / (7.5 x 10^14) = 300,000,000 / 750,000,000,000,000 = 0.0000004
Hope this helps :)
</span>
6 0
3 years ago
Read 2 more answers
HEY CAN ANYONE PLS PLS ANSWER DIS I RLY NEED IT
jeka94
Place the object in an electronic balance and measure its mass.
Place a measured amount of water in the cylinder.
Place the object in the cylinder so that it’s fully submerged.
Measure the new level of the liquid and subtract the original level. This is equal to the volume of the object.

Density = mass / volume.
8 0
3 years ago
A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe
Archy [21]

Answer:

a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

Explanation:

a. Its angular speed in radians per second  ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

  = (1410 - 207)/(80.5/60)

  = 60(1410 - 207)/80.5

  = 60(1203)80.5

  = 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and  t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

  = 207 × 1.342 + 1/2 × 896.65 × 1.342²

  = 277.725 + 807.417

  = 1085.14 revolutions ≅ 1085 revolutions

5 0
3 years ago
weegy a 7.5kg block is placed on a table. if its bottom surface area is 0.6m2 , how much pressure does the block exert on the ta
Lesechka [4]

The pressure exerted by the block on the table is given by:

p=\frac{W}{A}

where W is the weight of the box, and A is the bottom surface area of the box.

The weight of the box is: W=mg=(7.5 kg)(9.81 m/s^2)=73.6 N

Substituting into the first equation, we find the pressure:

p=\frac{W}{A}=\frac{73.6 N}{0.6 m^2}=122.7 Pa

4 0
3 years ago
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