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3 years ago

The pressure in a water line is 1500 kpa. what is the line pressure in (a) lb/ft2 units and (b) lbf/in2 (psi) units?

Physics

2 answers:

kotegsom [21]3 years ago

7 0

To solve this problem, we know that:

1 psi = 6894.76 Pa

1 lb / ft^2 = 47.88 Pa

Therefore:

a. 1500 x 10^3 Pa * (1 lb / ft^2 / 47.88 Pa) = 31,328.32 lb / ft^2

b. 1500 x 10^3 Pa * (1 psi / 6894.76 Pa) = 217.56 psi

Vladimir [108]3 years ago

4 0

Answer:

a) 1500 kpa = 31,328.3208 lb/ft2

b) 1500 kpa = 217.5 psi

Explanation:

This question can be solved using proportionality, that is, simple rules of three.

We have that:

Pressure of water line: 1500 kpa

(a) lb/ft2 units?

Each lb/ft2 is 0.04788 kpa. So

1 lb/ft2 - 0.04788 kpa

x lb/ft2 - 1500 kpa

$0.04788x = 1500$

$x = \frac{1500}{0.04788}$

$x = 31328.3208$

1500 kpa = 31,328.3208 lb/ft2

(b) lbf/in2 (psi) units?

Each kpa is 0.145psi. So

1 kpa - 0.145 psi

1500 kpa - x psi

$x = 1500*0.145$

$x = 217.5$

1500 kpa = 217.5 psi

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Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8

sergejj [24]

Answer:

The hottest temperature is $T_2 = 39^o C$

Explanation:

From the question we are given

$T_1 = 98 F$

$T_2 = 39^oC$

$T_3 = 303 \ K$

Generally converting $T_3$ to Fahrenheit

$T_3' = (T_3 -273 ) * \frac{9}{5} + 32$

=> $T_3' = (303 -273 ) * \frac{9}{5} + 32$

=> $T_3' = 86 F$

Converting $T_2$ to Fahrenheit

$T_2' = T_2 * \frac{9}{5} + 32$

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Now comparing the temperature in Fahrenheit we see that $T_2$ is the hottest

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3 years ago

A man paddles a canoe at 6 km per hour. If he paddles on a river with a current of 6 km per hour, what is the speed of the canoe

Romashka-Z-Leto [24]

Answer:

If the canoe heads upstream the speed is zero. And directly across the river is 8.48 [km/h] towards southeast

Explanation:

When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:

Vr = velocity of the river = 6[km/h}

Vc = velocity of the canoe = -6 [km/h]

We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.

Vt = Vr + Vc = 6 - 6 = 0 [km/h]

For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.

So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).

Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:

$Vt = \sqrt{(6)^{2} +(-6)^{2} } \\Vt=8.48[km/h]$