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Nataliya [291]
3 years ago
15

Calculate the decrease in gravitational potential energy of the ball as it moves down through the 0.90m the mass of the ball is

0.60kg
Physics
1 answer:
icang [17]3 years ago
8 0

Answer:

G.P.E = 5.292 Joules

Explanation:

Given the following data;

Mass = 0.6 kg

Height = 0.9 m

To find the gravitational potential energy;

Gravitational potential energy (GPE): it is an energy possessed by an object or body due to its position above the earth.

Mathematically, potential energy is given by the formula;

G.P.E = mgh

Where,

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

But we know that acceleration due to gravity is equal to 9.8m/s²

G.P.E = 0.6*9.8*0.9

G.P.E = 5.292 Joules

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please tell th answer fast suppose A ball of mass M is thrown vertically upward with initial speed be its speed is continuously
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Answer:.

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Explanation:

The answer I think

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2 years ago
When a sacred item or symbol is removed from its special place or is duplicated in mass quantities, then it becomes profane as a
Galina-37 [17]

Answer:

It becomes profane as a result of desacralization

Explanation:

Because desacralization means when a dedicated religious structure is no longer used for its intended purpose both rather used for another purpose other than the original purpose

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2 years ago
Which kind of pigment reflects two primary light colors and absorbs one?1) primary pigment2) secondary pigment3) complementary p
weqwewe [10]

The primary colors of light are red, blue and green.

There are the pigments like yellow, magenta and cyan that are the mixture of two primary colors.

For example, magenta is a mixture of red and blue color. Thus, it reflects the red and blue color. Also, magneta absorbs the green color.

These type of colors that reflects two primary colors and absorb one color are known as secondary pigments.

Hence, 2nd option is the correct answer.

8 0
9 months ago
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user100 [1]
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3 0
3 years ago
Read 2 more answers
Consider a semicircular ring of radius R. Its linear mass density varies as lambda =lambda not sin theta. Locate its centre of m
bearhunter [10]

Answer:

(0, πR/4)

Explanation:

The linear mass density (mass per length) is λ = λ₀ sin θ.

A short segment of arc length is ds = R dθ.

The mass of this short length is:

dm = λ ds

dm = (λ₀ sin θ) (R dθ)

dm = R λ₀ sin θ dθ

The x coordinate of the center of mass is:

X = ∫ x dm / ∫ dm

X = ∫₀ᵖ (R cos θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

X = R ∫₀ᵖ sin θ cos θ dθ / ∫₀ᵖ sin θ dθ

X = R ∫₀ᵖ ½ sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R ∫₀ᵖ 2 sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R (-cos 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

X = ¼R (-cos 2π − (-cos 0)) / (-cos π − (-cos 0))

X = ¼R (-1 + 1) / (1 + 1)

X = 0

The y coordinate of the center of mass is:

Y = ∫ y dm / ∫ dm

Y = ∫₀ᵖ (R sin θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

Y = R ∫₀ᵖ sin² θ dθ / ∫₀ᵖ sin θ dθ

Y = R ∫₀ᵖ ½ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R ∫₀ᵖ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R (θ − ½ sin 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

Y = ½R [(π − ½ sin 2π) − (0 − ½ sin 0)] / (-cos π − (-cos 0))

Y = ½R (π − 0) / (1 + 1)

Y = ¼πR

4 0
3 years ago
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