Answer:
A car would need to be moving with a velocity of 28.86 m/s for its acceleration to be numerically equal to the acceleration due to gravity.
Explanation:
Radial acceleration for an object in circular motion is given by
α = v²/r
α = g = 9.8 m/s²
v = ?
r = 85 m
9.8 = v²/85
v² = 9.8 × 85 = 833
v = 28.86 m/s
Everything would be more advanced and logical while the past people really just tried to survive.
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:
475 m , 950 m
Explanation:
Let l be the length of the side perpendicular to the barn.
1900-2l = length of the side parallel to the barn
Area A= l( 1900-2l)
A= 1900l-2l^2
now, the maximum value of l ( the equation being quadratic)
l_max= -b/2a
a= 2
b=1900
l_max= -1900/4= 475 m
then 1900-2l= 1900-2×(475) = 950 m
So, the dimensions that maximize area are
950 and 475
Now. A_max = -2( l_max)^2+1900×l_max
A_max= -2(475)^2+1900×475
A_max= 451250 m^2
or, 475×950 = 451250 m^2