4. A gas's solubility is best in a liquid solvent when the solution is under high or low pressure

Floor lamps usually have a base with large inertia, while the long body and top have much less inertia. Part A If you want to sh

Novay_Z [31]

Answer:

When a an object is been rotated its resistance capacity to that rotational force is know as rotational inertia and this mathematically given as

$I = mr^2$

Where m is the mass

r is the rotation radius

For the spinning of the lamp as a baton to work the location of the center of mass of the floor lamp needs to be located

This is more likely to be located closer to base of the lamp as compared to the top, so success of spinning a floor lamp like a baton is highly likely if the lamp is grabbed closer to the base because that is where the position of its center of mass is likely to be.

Explanation:

5 0

3 years ago

If a jet travels 350 m/s, how far will it travel each second?

77julia77 [94]

Answer:

It will travel 350 meters each second.

Explanation:

The unit rate, 350 m/s, tells us that the jet will travel 350 meters per every second elapsed.

6 0

3 years ago

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A charge q moves from point A to point B. The potential energy of the charge at point A is 5.3 _ 10-10 joules and at point is B

stiv31 [10]

The change in electric potential energy will be converted to kinetic energy; thus:
K.E = 5.3 x 10⁻¹⁰ - 3.1 x 10⁻¹⁰
K.E = 2.2 x 10⁻¹⁰ Joules
Option D is correct.

7 0

3 years ago

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A fireman standing on a 15 m high ladder operates a water hose with a round nozzle of diameter 2.02 inch. The lower end of the h

Nataly_w [17]

Answer:

v₁ = 1,606 10⁴ m / s

Explanation:

For this exercise we must use Bernoulli's equation, let's use index 1 for the nozzle on the stairs and index 2 the pump on the street

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

The pressure when the water comes out is the atmospheric pressure

P₁ = P_atm

The difference in height between the street and the nozzle on the stairs is

y₂-y₁ = 15 m

Now let's use the continuity equation

v₁ A₁ = v₂ A₁

The area of a circle is

A = π r² = π (d/2)²

v₁ π d₁²/ 4 = v₂ π d₂²/ 4

v₂ = v₁ d₁² / d₂²

Let's replace

P₂-P_atm + ½ ρ [ (v₁ d₁² / d₂²)²- v₁² ] + ρ g (y2-y1) = 0

P₂- P_Atm + ρ g (y₂-y₁) = ½ ρ v₁² [1- (d₁/d₂)⁴]

v₁² [1- (d₁/d₂)⁴] = (P₂-P_atm) ρ / 2 + g (y₂-y₁) / 2

Let's reduce the magnitudes to the SI system

d₂ = 3.37 in (2.54 10⁻² m / 1 in) = 8.56 10⁻² m

d₁ = 2.02 in = 5.13 10⁻² m

Let's calculate

v₁² [1- (5.13 / 8.56) 4] = 449.538 10³ 10³/2 -9.8 15/2

v₁² [0.8710] = 2.2477 10⁸ - 73.5 = 2.2477 10⁸

v₁ = √ 2.2477 10⁸ /0.8710

v₁ = 1,606 10⁴ m / s

4 0

3 years ago

Can someone help me please

nlexa [21]

Answer:

b

Explanation:

8 0

2 years ago

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