All categories

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

You might be interested in

A force of 10 n acts on a mass of 5 kg producing an acceleration of 1.5 m/s2 what is the magnitude of the force of friction acti

prisoha [69]

The friction is 2.5N. The Net force is 10 N - 2.5 N .= 7.5 N.
acceleration = 7.5 / 5 = 1.5 m/s^2

8 0

3 years ago

1. The geologic time scale divides time into years and centuries, true or false

hodyreva [135]

Answer:

The answer is "False"

Explanation:

The geologic time scale is the "schedule" for occasions in Earth history. It partitions time into named units of unique time called in descending order of duration "eons, eras, periods, epochs, and ages". The specification of those geologic time units depends on stratigraphy, which is the relationship and order of rock layers. The fossil structures that happen in the stones, nonetheless, give the central methods for setting up a geologic time scale, with the circumstance of the development and vanishing of far and wide species from the fossil record being used to outline the beginnings and endings of ages,, periods, and different stretches.

Geologic time is the broad time period involved by the geologic history of Earth. Formal geologic time starts toward the beginning of the Archean Eon (4.0 billion to 2.5 billion years back) and proceeds to the current day.

5 0

3 years ago

a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec

MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

$\frac{1}{2} mu_1^2+0=\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2$

∴ $\frac{1}{2} m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2$

∴ $\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2$

∴ $mv$ ₁ $v$ ₂ = 0

It is impossible for the mass to be zero.
Because the second ball moves, velocity v2 cannot be zero.
As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.

<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

#SPJ4

3 0

2 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.60×106 N, one at an angle 13.0 west of north, an

Juliette [100K]

Answer: $W=1.93\times 10^9 J$