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Minchanka [31]
4 years ago
7

What is the horizontal component of the force on the ball after it leaves the throwers hands

Physics
1 answer:
e-lub [12.9K]4 years ago
5 0
After the ball leaves the thrower's hand, the only force on it is
due to gravity.  There's no horizontal force acting on it at all. (C) 
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Daltons theory was identified using
Margaret [11]
<h2>Answer with Explanation </h2>

Dalton’s theory can be classified by the following hypotheses:

1) All material was formed of particles, unbreakable and strong construction segments.

2) All particles of a given component are indistinguishable in volume and characteristics

3) Compounds are determined by a mixture of two or more distinct kinds of atoms.

4) Chemical responses appeared in the rearrangement of the reacting atoms.

This theory was to explain all matter in terms of atoms and their characteristics, the law of conservation of volume and the law of constant composition.

7 0
4 years ago
The barometric pressure in breckenridge, colorado (elevation 9600 feet) is 580 mm hg. how many atmospheres is this?
Maksim231197 [3]

1 atmospheric pressure = 760.0 mm Hg

Thus 580 mm Hg = (580 mm Hg/(760 mm Hg/atm))

= 0.763 atm


7 0
3 years ago
Maxwell shoots a rubber band at his friend Jimmy. Which type of energy is converted into kinetic energy?
Leokris [45]

A stretched rubber band is storing <em>elastic potential energy. (A)</em>

3 0
3 years ago
Water (rhoH20 = 1000.0 kg/m3 ) flows through a garden hose that goes up a step 20.0 cm high. The cross-sectional area of the hos
Soloha48 [4]

Answer:

 P₂ = 138.88 10³ Pa

Explanation:

This is a problem of fluid mechanics, we must use the continuity and Bernoulli equation

Let's start by looking for the top speed

        Q = A₁ v₁ = A₂ v₂

We will use index 1 for the lower part and index 2 for the upper part, let's look for the speed in the upper part (v2)

         v₂ = A₁ / A₂ v₁

They indicate that A₂ = ½ A₁ and give the speed at the bottom (v₁ = 1.20 m/s)

         v₂ = 2  1.20

         v₂ = 2.40 m / s

Now let's write the Bernoulli equation

        P₁ + ½ ρ v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂

Let's clear the pressure at point 2

       P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁-y₂)

we put our reference system at the lowest point

        y₁ - y₂ = -20 cm

Let's calculate

       P₂ = 143 10³ + ½ 1000 (1.20² - 2.40²) + 1000 9.8 (-0.200)

       P₂ = 143 103 - 2,160 103 - 1,960 103

       P₂ = 138.88 10³ Pa

3 0
4 years ago
Ways that charges can be transferred
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3 years ago
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