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3 years ago

A satellite orbits the Earth at constant speed. Is this an unbalanced or balanced force?

Physics

1 answer:

natulia [17]3 years ago

3 0

Answer:

An unbalanced force is one that causes a change in the motion of the object to which the force is applied. An object at rest or an object in steady motion continues at rest or in unchanged motion unless it is subjected to an unbalanced force.

Balance forces are two forces acting in opposite directions on an object, and equal in size. Anytime there is a balanced force on an object, the object stays still or continues moving continues to move at the same speed and in the same direction.

Explanation:

A satellite orbiting Earth at constant speed would be considered an unbalanced force because it is an object in steady motion and it will remain in steady motion unless it is subjected to an unbalanced force.

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A woman does 236 J of work

Ratling [72]

Answer:

The woman's force was directed 59.22⁰ to the horizontal.

Explanation:

Given;

work done by the woman, W = 236 J

distance through the load was moved, d = 24.4 m

applied force, F = 18.9 N

inclination of the force, = θ

The work done by the woman is calculated as;

W = Fdcosθ

$cos \ \theta = \frac{W}{Fd} \\\\cos \ \theta = \frac{236 }{18.9 \times 24.4} \\\\cos \ \theta = 0.5118\\\\\theta = cos^{-1} ( 0.5118)\\\\\theta =59.22^0$

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3 years ago

A large aquarium of height 6 m is filled with fresh water to a depth of D = 1.50 m. One wall of the aquarium consists of thick p

Vesnalui [34]

Answer

given,

height of aquarium = 6 m

Depth of fresh water = D = 1.50 m

horizontal length of the aquarium(w) = 8.40 m

total force increased when liquid is filled to depth = 4.30 m

g = 9.81 m/s²

ρ = 998 Kg/m³

force in the aquarium.

dF = PdA

$F = \int PdA$

$F = \int \rho\ g\ y\ (wdy)$

$F = \rho\ g\ w \int ydy$

$F = \rho\ g\ w\dfrac{y^2}{2}$

$F = \dfrac{\rho\ g\ w\ y^2}{2}$

At D = 1.5 m

$F = \dfrac{980\times 9.8\times 8.4\times 1.5^2}{2}$

F = 9.08 x 10⁴ N

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$F = \dfrac{980\times 9.8\times 8.4\times 4.3^2}{2}$

F = 7.46 x 10⁵ N

Total force on the wall increased by

ΔF = 74.6 x 10⁴ - 9.08 x 10⁴

ΔF = 65.52 x 10⁴ N

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