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sdas [7]
2 years ago
11

I need to know which ones are physical or chemical changes.

Physics
1 answer:
andrezito [222]2 years ago
7 0
Physical
Chemical
Physical
Chemical
Chemical
Chemical
Physical
Physical
Physical
Chemical
Chemical
Physical
Physical
Physical
Chemical
Chemical
Chemical
Chemical
Physical
Chemical
Chemical
Chemical
Chemical
Chemical
Chemical

I believe this is right but double check to make sure :)
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In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 268 kg and moves
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Before we go through the questions, we need to calculate and determine some values first.

r = 11.5 m 
<span>m = 280 kg </span>
<span>Centripetal force = m x v^2/r = 280 x (17.1^2/11.5) = 7119.55 N 
</span>
1) What is the magnitude of the normal force on the care when it is at the bottom of the circle. 

<span>Centripetal force + mg = 7119.55 + (280 x 9.8) = 9863.55 N </span>

<span>2) What is the magnitude of the normal force on the car when it is at the side of the circle. </span>

<span>Centripetal force = 7119.55 N </span>


<span>3) What is the magnitude of the normal force on the car when it is at the top of the circle. </span>

<span>Centripetal force - mg = 7119.55 - (280 x 9.8) = 4375.55 N </span>

<span>4) What is the minimum speed of the car so that it stays in contact with the track at the top of the loop. </span>

√<span>(gr) </span>

√<span>(9.8 x 11.5) = 10.62 m/s</span>
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What regions contain the largest concentration of oil wells
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A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of
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Answer:

a) P=2450\ Pa

b) \delta h=23.162\ cm

Explanation:

Given:

height of water in one arm of the u-tube, h_w=25\ cm=0.25\ m

a)

Gauge pressure at the water-mercury interface,:

P=\rho_w.g.h_w

we've the density of the water =1000\ kg.m^{-3}

P=1000\times 9.8\times 0.25

P=2450\ Pa

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Now the same pressure is balanced by the mercury column in the other arm of the tube:

\rho_w.g.h_w=\rho_m.g.h_m

1000\times 9.8\times 0.25=13600\times 9.8\times h_m

h_m=0.01838\ m=1.838\ cm

<u>Now the difference in the column is :</u>

\delta h=h_w-h_m

\delta h=25-1.838

\delta h=23.162\ cm

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