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rodikova [14]
2 years ago
10

6

Engineering
2 answers:
grin007 [14]2 years ago
8 0

Answer:

(1)speed at which ball B was thrown upward-13.95m/s (2) velocity pf a=14m/s and b=7.25 m/s

Explanation:

Eduardwww [97]2 years ago
7 0

Answer:

ball a is released from rest at a height of 20 m

Explanation:

You might be interested in
Which of the following are considered software piracy? Check all of the boxes that apply.
Serga [27]

Answer:

The answer is copying a program to give to someone else to use

and burning a copy of a DVD to sell

Explanation:

Software piracy is the act of stealing software that is legally protected. This stealing includes copying, distributing, modifying or selling

7 0
2 years ago
True or False: Stress can effectively be relieved through physical activity, getting enough rest and sleep, and relaxation techn
Neko [114]

Answer:

True

Explanation:

Actually this are some of the nitty gritty answers and ways to control or stip stress, lemme explain them you see stress most of the time may come from deep thoughts that are and are pushing you to the wall, and you in your state you as you react to that which is so demanding as it makes you tense,so if at you will incorporate physical activities like running, jogging or push ups the psychological tension in you is broken as you focus so much on the activities rather than the tension, then getting enough rest cools the mind and all your thoughts settle as in your brain starts to adapt to chilling and relaxation and enough sleep will actually make ones head to be at peace because if you lack enough sleep you might have an excruciating migraine when you are stressed up and finally relaxation techniques makes the body accept the situation and then you manouver out of it as you grow strongly.

Hope this will help!

7 0
2 years ago
Read 2 more answers
2. The moist weight of 0.1 ft3 of soil is 12.2 lb. If the moisture content is 12% and the specific gravity of soil solids is 2.7
adell [148]

The answers to dry unit weight, void ratio, porosity, degree of saturation, volume occupied by water are respectively;

γ_d = 108.93 lb/ft³; e = 0.56; n = 0.36; S = 0.58; V_w = 0.021 ft³

<h3>Calculation of Volume and Weight of soil</h3>

We are given;

Moist weight; W = 12.2 lb

Volume of moist soil; V = 0.1 ft³

moisture content; w = 12% = 0.12

Specific gravity of soil solids; G_s = 2.72

A) Formula for dry unit weight is;

γ_d = γ/(1 + w)

where γ_w is moist unit weight as;

γ_w = W/V

γ_w = 122/0.1 = 122 lb/ft³

Thus;

γ_d = 122/(1 + 0.12)

γ_d = 108.93 lb/ft³

B) Formula for void ratio is;

e = [(G_s * γ_w)/γ_d] - 1

e = [(2.72 * 122)/108.93] - 1

e = 0.56

C) Formula for porosity is;

n = e/(1 + e)

n = 0.56/(1 + 0.56)

n = 0.36

D) Formula for degree of saturation is;

S = (w * G_s)/e

S = (0.12 * 2.72)/0.56

S = 0.58

E) Volume occupied by water is gotten from;

V_w = S*V_v

where;

V_v is volume of voids = nV

V_v = 0.36*0.1

V_v = 0.036 ft³

Thus;

V_w = 0.58 * 0.036

V_w = 0.021 ft³

Read more about Specific Gravity of Soil at; brainly.com/question/14932758

4 0
2 years ago
Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve
Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature T_1 = -16^0\ C

Quality x_1 = 0.2

Outlet:

Temperature T_2 = -16^0 C

Quality  x_2 = 1

The following data were obtained at saturation properties of R134a at the temperature of -16° C

v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\  v_g = 0.1247 \ m^3 /kg

v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\  v_1 = 0.0255 \ m^3/kg \\ \\ \\  v_2 = v_g = 0.1247 \ m^3/kg

m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\  m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6  \\ \\ \mathbf{m = 0.0534 \ kg/sec}

\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2  \\ \\  \rho_1v_1 = \rho_2v_2   \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}

3 0
2 years ago
1. Sewage-treatment plant, a large concrete tank initially contains 440,000 liters liquid and 10,000 kg fine suspended solids. T
Elenna [48]

Answer:

Concentration = 10.33 kg/m³

Explanation:

We are given;

Mass of solids; 10,000 kg

Volume; V = 440,000 L = 440 m³

Rate at which water is pumped out = 40,000 liter/h

Thus, at the end of 5 hours we amount of water that has been replaced with fresh water is = 40,000 liter/h x 5 hours = 200,000 L = 200 m³

Now, since the tank is perfectly mixed, therefore we can calculate a ratio of fresh water to sewage water as;

200m³/440m³ = 5/11

Thus, the amount left will be calculated by multiplying that ratio by the amount of solids;

Thus,

Amount left; = 10000 x (5/11) = 4545 kg

The concentration would be calculated by:

Concentration = amount left/initial volume

Thus,

Concentration = 4545/440 = 10.3 kg/m³

8 0
3 years ago
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