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3 years ago

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

Physics

2 answers:

tatuchka [14]3 years ago

7 0

Cell phone carriers

dezoksy [38]3 years ago

6 0

Answer:

In the United States,cell carrier services are often considered to be a part of an oligopoly

Explanation:

Cell carrier services are a brand value and imagine

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What kind of device can you use to to separate visible light into its different colors

juin [17]

A prism will separate white light into a rainbow of light

5 0

3 years ago

Estimated speed of the vehicle

SSSSS [86.1K]

Really, Gundy ? ! ?

The formula for the car's speed is given and discussed in the box. The formula is

v = √(2·g·μ·d)

Then they tell you that μ is 0.750 , and then they tell you that d = 52.9 m . Also, everybody knows that 'g' is gravity = 9.8 m/s² .

They also tell us that the mass of the car is 1,000 kg, and they tell us that it took 3.8 seconds to skid to a stop. But we already have all the numbers in the formula without knowing the car's mass or how long it took to stop. The police don't need to weigh the car, and nobody was there to measure how long the car took to stop. All they need is the length of the skid mark, which they can measure, and they'll know how fast the guy was going when he hit the brakes !

Now, can you take the numbers and plug them into the formula ? ! ?

v = √(2·g·μ·d)

v = √( 2 · 9.8 m/s² · 0.75 · 52.9 m)

v = √( 777.63 m²/s²)

v = 27.886 m/s

Rounded to 3 digits, that's 27.9 m/s .

That's about 62.4 mile/hour .

3 0

4 years ago

You will not know this. Really I didn't. What is the darkest known substance.

djyliett [7]

It darkness or my depression who knows

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3 years ago

A gazelle leaps from a cliff 2.5 m high with a speed of 5.6m/s.

jeyben [28]

Initial speed of Gazelle is along x direction and its value will be

$v_x = 5.6 m/s$

also its initial height is given as

$y = 2.5 m$

Part a)

now from kinematics along Y direction

$\Delta y = v_y t + \frac{1}{2} at^2$

as we know that

$\Delta y = 0$

$v_y = 0$

$a = 9.8 m/s^2$

$2.5 = 0 + \frac{1}{2} (9.8) t^2$

$t = 0.714 s$

Part b)

distance moved horizontally

$\Delta x = v_x t$

as we know that

$v_x = 5.6 m/s$

now we will have

$v_x = 5.6 (0.714) = 4m$

so it will lend at distance of 4 m.

Part c)

final velocity in vertical direction

$v_{fy} = v_y + at$

$v_{fy} = 0 + (9.8)(0.714) = 7 m/s$

$v_x = 5.6 m/s$

so net speed will be

$v^2 = v_x^2 + v_y^2$

$v^2 = 7^2 + 5.6^2$

$v = 8.96 m/s$

7 0

3 years ago

Two identical conducting spheres each having a radius of 0.500 cm are connected by a light, 1.90-m-long conducting wire. A charg

Ivan

Answer: 4.19 N

Explanation: In order to determinate the tension applied on the wire we have to calculate the electric force between the conductor spheres connected by the wire.

As the wire is a conductor the spheres are at same potential so we have:

V1=V2

V1=k*Q1/r1 and V2=k*Q2/r2

where r1=r2, then

Q1=Q2

so the electric force is given by:

F=k*Q^2/d^2 where d is the distance between the spheres.

Finally replacing the values, we have

F=9*10^9(41*10^-6)^2/(1.9)^2= 4.19 N

8 0

3 years ago