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Tom [10]
3 years ago
6

Some miners wish to remove water from a mine shaft. A pipe is lowered to the water 90 m below, and a negative pressure is applie

d to raise the water. (a) Calculate the pressure needed to raise the water. (b) What is unreasonable about this pressure? (c) What is unreasonable about the premise?
Physics
1 answer:
WITCHER [35]3 years ago
4 0

Answer:

Explanation:

Given

Pipe is lowered to the water h=90\ m

Negative Pressure is applied to raise the water

Pressure is given by

P=\rho gh

where P=pressure

\rho =Density

h=depth

P=10^3\times 9.8\times 90

P=8.82\times 10^{5}\ N/m^2\approx 8.82\ atm

(b)8.82 atm is much lower than the vapor pressure of water

(c)The fact of applying a negative pressure of 8.74 below the vapor pressure of water

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The horizontal motion of a launched projectile affects its vertical motion.
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The horizontal motion of a launched projectile affects its vertical motion is false

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<u>Explanation: </u>

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8 0
3 years ago
A wrecking ball is hanging at rest from a crane when suddenly the cable breaks. The time it takes for the ball to fall halfway t
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Answer:

1.697s

Explanation:

We use the second equation of free fall under gravity as follows;

h=ut+\frac{1}{2}gt^2...............(1)

Since  the ball fell freely, u = 0m/s, therefore equation (1) reduces to

h=\frac{1}{2}gt^2...............(2)

Given that h is the total height the ball falls through in time t seconds.

However, according to the stated problem the ball falls halfway in 1.2s, this simply implies that the ball falls through a distance of \frac{h}{2} in 1.2s. Hence we can write the following, given that g=9.8m/s^2;

\frac{h}{2}=\frac{1}{2}*9.8*1.2^2\\hence\\h=9.8*1.2^2\\h=14.112m

We can now proceed to find the time t for which it falls through h = 14.112m as follows;

14.112=\frac{1}{2}*9.8*t^2\\14.112=4.9t^2\\t^2=\frac{14.112}{4.9}\\t^2=2.88\\t=\sqrt{2.88} \\t=1.697s

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