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Alona [7]
3 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]3 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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Someone please help with this
SSSSS [86.1K]

Answer:

<em>The new force is 2/3 of the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}

Factoring out 2/3:

\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}

Substituting the original force:

F'=\frac{2}{3}F

The new force is 2/3 of the original force

7 0
3 years ago
A triangular plate with a non-uniform areal density has a mass M=0.500 kg. It is suspended by a pivot at P and can oscillate as
malfutka [58]

The period of the oscillations.T = 1.2042s

Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.

Typically, the mathematical formula for the moment of inertia is

T = 2 π √(I / mgd)

Therefore, a moment of inertia

I = 9.00×10-3 + md^2 ;

I=9.00*10^{-3}+ 0.5 * 0.3^2

I=0.054

T=2\pi \sqrt{0.5*9.8*0.3}

T=1.2042s

The period of the oscillations.T = 1.2042s

Read more about the period of the oscillations. brainly.com/question/14394641

#SPJ1

6 0
1 year ago
What exited kaveh pahlevan about his work?
USPshnik [31]

Answer:

i dont know

Explanation:

i dont know since you didn't provide something to base off of

7 0
2 years ago
Sound waves are also called compression waves. This means that as the wave travels through air, the ____ in creases and decrease
Alja [10]
The speed creases and decreases
5 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
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