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kogti [31]
3 years ago
9

A volcanic eruption launches two rocks of different masses into the air. The first rock had a mass of 10 kilograms while the sec

ond rock had a mass of 20 kilograms.
At the eruption, both rocks accelerated at a rate of 600 m/s2. Which of the following statements is true regarding the eruption of the two rocks?
The amount of force applied to rock one was twice as much as the force applied to rock two during the eruption.
The amount of force applied to rock one was the same as the force applied to rock two during the eruption.
The amount of force applied to rock one was half the amount of force applied to rock two during the eruption.
The amount of force applied to the rocks is impossible to determine without knowing the inertia of the rocks.
Physics
2 answers:
pashok25 [27]3 years ago
8 0

Answer:

C

Explanation:

just took test

Artist 52 [7]3 years ago
3 0

Answer:

C. The amount of force applied to rock one was half the amount of force applied to rock two during the eruption.

Explanation:

A force is an agent which changes the state of an object well applied to it. Second Newton's law of motion state's that;

F = ma

where F is the force, m is the mass of the object and a is the acceleration of the object.

In the given question,

Mass of first rock = 10 kilograms

Mass of the second rock = 20 kilograms

Acceleration of the two rocks = 600 m/s^{2}

Thus,

Force on the first rock = ma

                                    = 10 x 600

                                    = 6000 N

Force on the second rock = 20 x 600

                                   = 12000 N

Therefore, it can be observed that the amount of force applied to rock one was half the amount of force applied to rock two during the eruption.

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Một lò xo có cấu tạo đồng đều, chiều dài l0 và độ cứng k0 = 30 N/m. Cắt lò xo thành hai phần có chiều dài tương ứng l1 và l2, độ
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After the big bang, atoms in gas clouds experienced a greater gravitational pull to each other than atoms in other regions of th
allsm [11]
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These are the two statements with scientific facts that explain the described phenomenon
<span>
Gravitation between two objects increases when the distance between them decreases.</span>

When the mass of an object increases, its gravitational pull also increases.

Justification:

Those two facts are represented in the Universal Law of Gravity discovered by the scientific Sir Isaac Newton (1642 to 1727) and published in his book <span>Philosophiae naturalis principia mathematica.</span>

That law is represented by the equation:

F = G × m₁ × m₂ / d²

The product of the two masses on the numerator accounts for the fact that the gravitational force is directly proportional to the product of the masses, which is that as the masses increase the attraction also increase.

The term d² (square of the distance that separates the objects) in the denominator accounts for the fact that the gravitational force is inversely proportional to the square of the distance; that is as the separation of the objects increase the gravitational force decrease.


6 0
3 years ago
A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of rho0 is placed in a container of water. Initially
ivanzaharov [21]

Answer:

a) s,f,r  b) r c) f

Explanation:

To determine what happens with the sphere we use Newton's second law with the Archimedes principle that states that the thrust (B) on a body is equal to the weight of the liquid dislodged

For the sphere to be in equilibrium the sum of forces is zero

    B - W = 0

    B = W = mg

Now let's use the concept of density for the body and water

Solid sphere

   ρ = m / V

  V = 4/3 π r³

   m = ρ₀ (4/3 π r³)

   W = ρ₀ (4/3 π r³) g

Water  (a)

   ρ = mₐ / Vₐ

   mₐ = ρ Vₐ

   B = ρ Vₐ g

Let's replace and simplify

   ρ Vₐ g = ρ₀ (4/3 π r³) g

    ρ Vf = ρ₀ (4/3 π r³)            (1)

For the initial condition with rho, mo and ro the height of the water is H, let's analyze each case

a) We have the same mass, but less radius, as density is mass over volume density increases

   r  <ro        V <V₀   ⇒      ρ₁> ρ₀

When analyzing the equation (1) on the right side, this case is the most complicated because I can make the relationship between the density of the sphere and its volume change even when the mass is constant

Assume the three possibilities

- The product of (ρ₁ V) that does not matter in that case the left side does not change and the mark remains the same (s)

- The product (ρ₁ V) increases the left side must increase so the mark goes up (r)

- The product (ρ₁ V) decreases the left side should go down, so the low mark (f)

b) sphere the same radius, but the density increases.

In this case the right side of the equation (1) increases, therefore the left side must increase so that the volume must increase and consequently increase the height (r)

c) you have the same radius, but the mass decreases

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The right side of the equation decreases, because the density decreases, the left side must decrease, for this the volume must decrease, lowering the height (f)

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