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algol13
3 years ago
10

Wind resistance ___as speed increase.

Physics
1 answer:
chubhunter [2.5K]3 years ago
6 0

Answer:

increases dramatically

Explanation:

due to resistance tends to prevent the motion of a body. When speed increases, the particles of air are set into motion and this energizes them. More air gains increased acceleration and they collide with a body. This initiates a resistance against the motion of such body. for instance, wind resistance increases rapidly as speed increases.

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First you will want to sketch out both of the situations. It should be two sketches, one for the flagpole and one for the building.

To solve this, you will want to create a proportion.

Flagpole height/flagpole shadow=building height/building shadow

Therefore, it should look like this:
50/30= 300/x

Solve for x:
50x=9,000
X= 180 feet

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1. Calculate the work done by a 47 N force pushing a pencil 0.26 m.<br> 47x .26 = 12J<br> Solve it
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3 years ago
A 1.6-lb collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an
romanna [79]

Answer:

k = 652 lb/ft

Explanation:

Given :

Weight of the collar = 1.6 lb

The upstretched length of the spring = 6 in

Speed  = 16 ft/s

PA = 8 + 10

     = 18 inch

Let the initial elongation be $\Delta x_i$

∴ $\Delta x_i$ = 18 - 6

         = 12 inch = 1 foot

$PB = \sqrt{13^2+5^2}$

      = 13.925 inch

Final elongation in the spring

$\Delta x_B = 7.928 $ inch = 0.66 feet

Applying the conservation of the mechanical energy between A and B is

$K.E_A+P.E_{g,A}+P.E_{sp,A}= K.E_B+P.E_{g,B}+P.E_{sp,B} $

$0+mg_r+\frac{1}{2}k(\Delta x_i)^2=\frac{1}{2}mv_B^2+0+\frac{1}{2}k(\Delta x_B)^2$

$\frac{1}{2}k[(1)^2-(0.66)^2]=\frac{1.6}{2}\times (16)^2-1.6 \times 32 \times \frac{5}{12}$

0.281 \ k =204.8-21.33

k = 652 lb/ft

5 0
3 years ago
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