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Alona [7]
2 years ago
8

When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of

the air. Model a cyclist as having cross-section area 0.40 m2 and, because the human body is not aerodynamically shaped, a drag coefficient of 0.90.
What is the cyclist's power output while riding at a steady 7.3 m/s?
Physics
1 answer:
ratelena [41]2 years ago
8 0

Answer:

Power output = 96.506 watts

Explanation:

Drag coefficient (Cd) = 0.9

V = 7.3 m/s

Air density (ρ) = 1.225 kg/m^(3)

Area (A) = 0.45 m^2

Let's find the drag force ;

Fd=(1/2)(Cd)(ρ)(A)(v^(2))

So Fd = (1/2)(0.9)(1.225)(0.45)(7.3^(2)) = 13.22N

Drag power = Drag Force x Drag velocity.

Thus drag power, = 13.22 x 7.3 = 96.506 watts

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Electromagnetic radiation is emitted when a charged particle moves through a medium faster than the local speed of light. This r
alexandr1967 [171]

Answer:

to create the particle the speed must be greater than 2.25 10⁸ m / s

Explanation:

In this exercise we must use the relation of the index of refraction with the speed of light in a vacuum and a material medium

           n = c / v

where c is the speed of light in the vacuum, v the speed of light in the material medium and n the ratio of rafraccio

in this case they give us that the medium matter water them that has a refractive index of

              n = 1,333

we clear

          v = c / n

let's calculate

           v = 3 10⁸ / 1,333

           v = 2.25 10⁸ m / s

to create the particle the speed must be greater than 2.25 10⁸ m / s

6 0
2 years ago
If the distance between us and a star is doubled, with everything else remaining the same, the luminosity Group of answer choice
Savatey [412]

Answer:

remains the same, but the apparent brightness is decreased by a factor of four.

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).

The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).

The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.

The apparent brightness of a star is measured in watts per square meter.

If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).

Some of the examples of stars are;

- Canopus.

- Sun (closest to the Earth)

- Betelgeuse.

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8 0
2 years ago
If Siobhan hits a 0.25 kg volleyball with 0.5 N of force, what is the acceleration of the ball?
Alekssandra [29.7K]

Answer:

2 meters per second²

Explanation:

8 0
2 years ago
Read 2 more answers
Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a
Anton [14]

Answer:

Explanation:

moment of inertia of each blade which is similar to rod rotating about its one end

= 1/3 ml²

moment of inertia of 3 blades = ml²

= 5500 x 46²

I = 11638 x 10³ kg m²

angular velocity = 2πn where n is rotation per second

n = 11 / 60

angular velocity = 2π x 11/60

= 1.1513 rad /s

angular momentum

= moment of inertia x angular velocity

=  11638 x 10³ x 1.1513

= 13399 x 10³ kg m² per second.

4 0
2 years ago
The Deligne Dam on the Cayley River is built so that the wall facing the water is shaped like the region above the curve y=0.3x2
ki77a [65]

Answer:

 F = 1.65 10⁸ N

Explanation:

In this pressure problem we have to use the definition of pressure

             P = dF / dA

            dF = P dA

we already have the expression for force and the pressure in a liquid is

            P = Po + rho g (H-y)

Where Po is the atmospheric pressure acting on both sides of the dam, whereby its contribution is canceled and (H-y) is the distance from the surface

Let's look for an expression for the area differential

            A = xy

            dA = dx dy

            y = 0.3 x²

            x = √(y/0.3)

Let's build our equation with these expressions and integrate between the initial limit where the height is measured from the bottom of the dam y = 0, x = 0 to the upper limit, let's call it H = 200m, x = RA y / 0.3 and F = 0

           ∫ dF = ∫ (ρ g (H-y) dx dy

           -F = ρ g [∫∫ H dy dx - ∫∫ ydy dx]

           F = ρ g [∫ H x dy - ∫ x2√2 / 2 dy

Let's evaluate between the limits of integration

           F = ρ g [∫ (H (√y /√0.3) dy - ∫ y/0.3 1/2 dy

           F = ρ g (H /√0.3 ∫ √y dy - 1 /0.6 ∫ y dy)

Let's do the second integral

           F = ρ g (H/√0.3 y^{3/2}) 2/3 - 1/0.6  y2 / 2)

           F = ρ g (2H/3√0.3 y^{3/2})  - 1/1.2  y2 )

We evaluate at the limits

          Y = 0

          Y = 38 m

         

          F = 1000 9.8 (2 200/3√3  √38³  - 1/1.2 38²)

          F = 9800 (18032 - 1203)

          F = 1.65 10⁸ N

5 0
2 years ago
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