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lorasvet [3.4K]
3 years ago
12

Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is r

eleased
Physics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

<em>171.5m</em>

Explanation:

The velocity of sound in water = 343m/s

Time taken = 1.00secs

using the formula to calculate the distance

2x = vt

x is the distance

v is the speed of sound

t is the time

x = vt/2

x = 343(1)/2

x = 171.5m

<em>hence their separation 1.00 s after the second object is released is 171.5m</em>

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5.0 atm

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Work and Power Practice Calculations
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Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo
yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour                    

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We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is   = 102^{\circ}-65.3^{\circ}=36.7^{\circ}

Speed of first plane  = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, r^2 representing the distance between the planes, we see that:

r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295

r = 741.6959 m

3 0
3 years ago
Find the y-component of this
Alborosie

Answer:

-0.0789 m

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Notice it is negative since it is below the x-axis.

4 0
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