Use the diagram to determine the voltage drop of the conductors. (Round the FINAL answer to two decimal places.)

A 3.52 kg steel ball is tossed upward from a height of 6.93 meters above the floor with a vertical velocity of 2.99 m/s. What is

Dafna1 [17]

Answer : The final velocity of the ball is, 12.03 m/s

Explanation :

By the 3rd equation of motion,

$v^2-u^2=2as$

where,

s = distance covered by the object = 6.93 m

u = initial velocity = 2.99 m/s

v = final velocity = ?

a = acceleration = $9.8m/s^2$

Now put all the given values in the above equation, we get the final velocity of the ball.

$v^2-(2.99m/s)^2=2\times (9.8m/s^2)\times (6.93m)$

$v=12.03m/s$

Thus, the final velocity of the ball is, 12.03 m/s

7 0

3 years ago

According to fire regulations in a town, the pressure drop in a commercial steel, horizontal pipe must not exceed 2.0 psi per 25

bonufazy [111]

Answer:

6.37 inch

Explanation:

Thinking process:

We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.

To determine the pressure drop in the pipe:

Using the Bernoulli equation for mass conservation:

$\frac{P1}{\rho } + \frac{v_{2} }{2g} +z_{1} = \frac{P2}{\rho } + \frac{v2^{2} }{2g} + z_{2} + f\frac{l}{D} \frac{v^{2} }{2g}$

thus

$\frac{P1-P2}{\rho } = f\frac{l}{D} \frac{v^{2} }{2g}$

The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.

Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F

from the tables

Re= 2.01 × 10⁵

Hence, f = 0.018

Therefore, pressure drop, (P1-P2)/p = 2.70 ft

This occurs at ae presure change of 1.17 psi

Correlating with the chart, we find that the diameter will be D= 0.513

= <u>6.37 in Ans</u>

7 0

3 years ago

You want to see both Michael and Meet Wally Sparks. If you purchase tickets for both

Aleonysh [2.5K]

$2.

Both tickets cost $1.50
$1.50 x 2 = $3
$5 - $3 = $2

6 0

2 years ago

A steel bolt has a modulus of 207 GPa. It holds two rigid plates together at a high temperature under conditions where the creep

VikaD [51]

Answer:

14.36((14MPa) approximately

Explanation:

In this question, we are asked to calculate the stress tightened in a bolt to a stress of 69MPa.

Please check attachment for complete solution and step by step explanation

7 0

3 years ago

A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri

babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are 73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

$T_{BE}-W=0$ hence

$T_{BE}=W=20*9.81=196.2 N$

Therefore, tension in the cable, $T_{BE}=196.2 N$

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

$196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0$

$24.525-39.24+0.2D_x=0$

$D_x=73.575 N$

Similarly,

$A_x-D_y=0$

$A_x=73.575 N$

Therefore, both reactions at A and D are 73.575 N

7 0

3 years ago