In the United States, which type of industry is often considered part of an oligopoly?

Please answer of this question

Bogdan [553]

Answer:

$\frac{\pi }{15}$ m

Explanation:

At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.

Angle between the two hands = 2/12 * 360

= 60°

Arc Length = $2\pi r(\frac{60}{360} )$

= $2\pi (0.2)(\frac{1}{6} )\\= \frac{\pi }{15} m$

8 0

2 years ago

At a fixed depth within a fluid at rest, the pressure pushing upward is

earnstyle [38]

Answer: E) A) salt water.

Explanation:

E) In equilibrium, pressure exerts equally in all directions, so for a given depth, the pressure is the same for all points located at the same depth, and it can be written as follows:

p = p₀ + ρ.g.h, where p₀ = atmospheric pressure, ρ=fluid density, h=depth from the surface.

A) The buoyant force, as discovered by Archimedes, is an upward force, that opposes to the weight of an object (as it is always downward), and is equal to the weight of the volume of the liquid that the object removes, which means that is proportional to the density of the liquid.

As salt water is denser than fresh water, the buoyant force exerted by the salt water is always greater than the one produced by the fresh water, so objects will float more easily in salt water than in fresh water.

In the limit, it is possible that one object float in salt water and sink in fresh water.

3 0

3 years ago

Please awnser and show the ways

topjm [15]

Answer:

Answers in solutions.

Explanation:

Question 6:

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

The density of the substance in Crown A;

Density = mass ÷ volume = $\frac{1930}{100}$ = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043 g/cm³ ≈ 10.5 g/cm³ (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3>The density of the mixture:</h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to $\frac{29.8}{2}$ = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

Question 7:

An empty masuring cylinder has a mass of 500 g.

Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.

The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷ volume = 350 g ÷ 100 cm³ = 3.5g/cm³

Question 8:

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

$\frac{0.02 * 1}{0.001}$ = 20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

$\frac{0.02 * 1,000,000}{1}$ = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

Question 9:

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

Question 10:

A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
in a displacement can
The mass of liquid which overflow is 20 g

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; $\frac{20 * 1}{1}$ = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density × volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

7 0

3 years ago

a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the

erma4kov [3.2K]

The emerging velocity of the bullet is 71 m/s.

The bullet of mass m moving with a velocity u has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed v when it emerges from the block.

If the block exerts a resistive force F on the bullet and the thickness of the block is x then, the work done by the resistive force is given by,

$W=Fx$