Please help me! i cant find this one anywhere!
1 answer:
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Answer:
0·95
Explanation:
Given the combined mass of the rider and the bike = 100 kg
Percent slope = 12%
∴ Slope = 0·12
Terminal speed = 15 m/s
Frontal area = 0·9 m²
Let the slope angle be β
tanβ = 0·12
As it attains the terminal speed, the forces acting on the combined rider and the bike must be balanced and therefore the rider must be moving download as the directions of one of the component of weight and drag force will be in opposite directions
The other component of weight will get balance by the normal reaction and you can see the figure which is in the file attached
From the diagram m × g × sinβ = drag force
Drag force = 0·5 × d × × v² × A
where d is the density of the fluid through which it flows
is the drag coefficient
v is the speed of the object relative to the fluid
A is the cross sectional area
As tanβ = 0·12
∴ sinβ = 0·119
Let the fluid in this case be air and density of air d = 1·21 kg/m³
m × g × sinβ = 0·5 × d × × v² × A
100 × 9·8 ×0·119 = 0·5 × 1·21 × × 15² × 0·9
∴ ≈ 0·95
∴ Drag coefficient is approximately 0·95
Answer:
<em>0.228 nm</em>
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Explanation:
Atomic mass number of copper = 64
but an atomic mass unit = 1.66 x 10^-27 kg
therefore, the mass of the copper atom m = 64 x 1.66 x 10^-27 kg = 1.06 x 10^-25 kg
The number of atoms in this mass n = ρ/m
where ρ is the density of copper = 8.96 x 10^3 kg/m^3
==> n = (8.96 x 10^3)/(1.06 x 10^-25) = 8.45 x 10^28 atoms/m^3
We know that the volume occupied by this amount of atoms n =
where a is the lattice constant
equating, we have
8.45 x 10^28 =
a = 4.389 x 10^9
we also know that
d = 1/a
where d is the smallest distance between the two copper atom.
d = 1/(4.389 x 10^9) = 2.28 x 10^-10 m
==> <em>0.228 nm</em>