A rider on a bike with the combined mass of 100kgattains aterminal speed of 15m/son a 12% slope. Assuming that the only forces affecting thespeed are the weight and the drag, calculate the drag coefficient. The frontal area is0.9m2.

C_d = 0.942

Explanation:

Let's first calculate the angle of inclination.

Formula is;

tan θ = (%slope)

% Slope is given as 12%

Thus;

θ = tan^(-1) (12/100)

θ = 6.843°

Let's now calculate the force due to the weight of the rider and bike combined from;

F = mg sin θ

We are given; m = 100 kg.

Thus;

F = 100 × 9.81 × sin 6.843

F = 116.885 N

The drag force will also be the same as the force due to the weight of the body. Thus;

Drag force; F = C_d(½ρu²A)

Where;

C_d is drag coefficient

ρ is density

U is terminal speed

A is area

We are given;

A = 0.9 m²

U = 15 m/s

From online tables, density of air is approximately 1.225 kg/m³

Thus;

116.885 = C_d(½ × 1.225 × 15² × 0.9)

116.885 = 124.03125C_d

C_d = 116.885/124.03125

C_d = 0.942

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m