A person pushes a large 42.9 kg box at a constant velocity of 9 m/s across a horizontal floor for 3.8 s. Find the
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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:
0.57 m
Explanation:
First of all, we need to calculate the time it takes for the ball to cover the horizontal distance between the starting position and the crossbar. This can be done by analzying the horizontal motion only. In fact, the horizontal velocity is constant and it is
And the distance to cover is
d = 19 m
So the time taken is
Now we want to find how high the ball is at that time. The initial vertical velocity is
So the vertical position of the ball at time t is
where g = 9.8 m/s^2 is the acceleration of gravity. Substituting t = 2.04 s, we find
The crossbar height is 3.05 m, so the difference is
So the ball passes 0.57 m above the crossbar.