Based on Archimedes' principle, the greatest buoyant force an object can experience in water is determined by which quantity?

Licemer1 [7]

Light travels in straight lines. Once a light has been produced, it will keep moving in a straight line until it hits something else. Shadows are evidence of light traveling in straight lines. An object blocks light so that it can’t reach the surface where we see the shadow.

8 0

3 years ago

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How does the ocean influence global climate?

Korvikt [17]

By absorbing solar radiation and releasing heat needed to drive the atmospheric circulation

3 0

3 years ago

The force of gravity on a 2-kg rock is twice as great as that on a 1-kg rock. why then doesn't the heavier rock fall faster?

katrin2010 [14]

It takes twice the force to produce the same acceleration in the 2kg rock.

5 0

3 years ago

Which factors affect gravitational force? Check all that apply.

alina1380 [7]

Answer:

distance between the objects

masses of the objects

Explanation:

Gravitational force is a force of attraction that pulls two objects with masses together.

To best understand the concept of gravitational force, newton's law of universal gravitation provides a good insight. The law states that "every object in the universe attracts each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".

From the law, we see that gravitational force is dependent on the masses of the object and their distances.

6 0

3 years ago

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Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=

drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

$v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2$

$v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2$

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

$v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

$v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

The final relative velocity of the satellite, $v_f$ = v₁ + v₂

∴ $v_f$ = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, $v_f$ = 0.190 m/s

4 0

3 years ago