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Papessa [141]
3 years ago
13

What would happen if an automobile moving at a high speed suddenly comes to a stop? How does this relate to Newton's laws and wh

at factors could change the effect of the sudden stop on the passengers?
Physics
1 answer:
insens350 [35]3 years ago
6 0
If an automobile moving at high speed suddenly comes to a stop, you would have a large change in momentum. This relates to Newton's second law in the form F = delta p / delta t, where p is momentum (mv).
You could lessen the effect of the sudden stop on the passengers by changing the average force exerted on them. If you look at Newton's second law again, you can see that given some delta p, you can decrease F by increasing delta t. What this means is that if you increase the length of time over which the change in momentum occurs, you can decrease the average force exerted to obtain that change in momentum. This is the reason why landing on a soft cushion is preferable to landing on a concrete surface. The cushion gives way to any object falling on it while still providing some resistance (you don't stop as abruptly), so while your change in momentum is the same in both cases, you have a larger delta t in the case of the cushion.
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notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

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Answer:

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