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Leni [432]
3 years ago
15

A weightlifter lifts a set of 1250kg weights a vertical distance of 2m weight lifting contest. what potential energy do the weig

hts now possess
Physics
1 answer:
kenny6666 [7]3 years ago
5 0

Answer: The potential energy of the weights is 24,500 J

Explanation:

For an object of mass M, that is at a distance H from the ground, the potential energy can be written as:

U = M*H*g

where g is the gravitational acceleration g = 9.8m/s^2

In this case, we have that:

mass of the weights = 1,250 kg

height = 2m

Then the potential energy of the weights is:

U = 2m*1,250kg*9.8m/s^2 = 24,500 J

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volume and erosion

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The compound magnesium phosphate has the chemical formula Mg3(PO4)2. In this compound, phosphorous and oxygen act together as on
S_A_V [24]

The question is incomplete, the complete question is;

The compound magnesium phosphate has the chemical formula Mg3(PO4)2. In this compound, phosphorus and oxygen act together as one charged particle, which is connected to magnesium, the other charged particle. What does the 2 mean in the formula 5Mg3(PO4)2? A. There are two elements in magnesium phosphate. B. There are two molecules of magnesium phosphate. C. There are two magnesium ions in a molecule of magnesium phosphate. D. There are two phosphate ions in a molecule of magnesium phosphate.

Answer:

There are two phosphate ions in a molecule of magnesium phosphate.

Explanation:

The compound magnesium phosphate is an ionic compound. Ionic compounds always consists of two ions, a positive ion and a negative ion.

In this case, the positive ion is Mg^2+ while the negative ion is PO4^3-.

The subscript, 2 after the formula of the phosphate ion means that there are two phosphate ions in each formula unit of magnesium phosphate.

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2 years ago
At a particular instant the magnitude of the momentum of a planet is 2.05 × 10^29 kg·m/s, and the force exerted on it by the sta
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3 years ago
What is Newton's 3rd Law of Motion? *
Alexandra [31]

Answer:

For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects.

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Every force sent into an object will sent a force in return.

If you smack a table, you can feel the table "push back"

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A batter hits a ball and it is caught 4 seconds later 100 m from home plate. What is the initial velocity (vector!!!) of the bal
Eddi Din [679]

Answer:

The initial velocity vector of the ball is;

\underset{u}{\rightarrow} = 25·i + 19.6·j

Explanation:

The given parameters are;

The time of flight of the ball = 4 seconds

The horizontal distance from the plate at which the ball is caught = 100 m

Let 'u', represent the initial velocity, we have;

u × cos(θ) × t = u × cos(θ) × 4 s = 100 m

u × cos(θ) = 25 m/s...(1)

t = \dfrac{2 \cdot u \cdot sin(\theta)}{g} = \dfrac{2 \cdot u \cdot sin(\theta)}{9.8 \ m/s^2} = 4 \ seconds

∴ 2·u·sin(θ) = 9.8 m/s² × 4 s = 39.2 m/s

\therefore \dfrac{2 \cdot u \cdot sin(\theta)}{u \cdot cos(\theta)} = 2 \cdot tan(\theta) = \dfrac{39.2 \ m/s}{25 \  m/s} = 1.568

tan(θ) = 1.568/2 = 0.784

θ = arctan(0.784) ≈ 38.096°

The direction of the velocity  vector of the ball, θ ≈ 38.096°

From equation (1), we have;

u × cos(θ) = 25 m/s

∴ u = 25 m/s/cos(θ) = 25 m/s/cos(arctan(0.784)) = 31.7672787629 m/s

The magnitude of the initial velocity vector, u = 31.7672787629 m/s

The vertical component of the initial velocity, u_y = u × sin(θ)

∴ u_y = 31.7672787629 × sin(arctan(0.784))

Therefore, the initial velocity vector of the ball is approximately, v = 31.767 m/s in a direction 38.096° above the horizontal, from which we have;

u = uₓ·i + u_y·j = 25·i + 19.6·j

\underset{u}{\rightarrow} = 25·i + 19.6·j

4 0
2 years ago
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