Which state of matter is most similar to solids

Answers

Answer 1

Answer:

Answer:

liquids

Explanation

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While a mason was working concrete into formwork, the formwork collapses. Who is BEST suites to rectify this problem? Mason Carpenter Project Manager O Construction Technician A device made in a workplace had defects. To address this issue the workshop manager should communicate directly with the workshop

Answers

Answer:

1. Carpenter

2. True

Explanation:

While a mason was working concrete into the formwork, the formwork collapses. The best person to rectify this problem is CARPENTER.

This is because it is the job of the Carpenter to design and build formwork, most especially wooden formwork. Formwork is like casing built to receive concrete and reinforcement during construction. Hence, when formwork collapses either due to stress, tension, or improper construction, it is the job of Carpenter to reconstruct the formwork or rectify the problem.

It is TRUE that when a device made in a workplace had defects. To address this issue the workshop manager should communicate directly with the workshop. However, this communication will be an instruction on what to do next, and it usually directs those responsible to take action where necessary. For example, a workshop manager communicates to a carpenter about the need to rectify a chair or table that has a defect.

A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the capacity of the highway section, the speed at capacity, and the density when the highway is at one-quarter of its capacity?

Answers

(a) The capacity will be "4006.4 veh/h".

(b) The speed at capacity be "25 mph".

(c) The density will be "299 veh/mi".

Given:

$q = 50k - 0.156 k^2$

At max. flow density,

$(dd)/(dk) =0$

$((dq)/(dt) ) = 50-0.321k =0$

(a)

→ $k = ((50)/(0.312) )$

$= 160.3 \ or \ 160 \ veh/mi$

By substituting the value,

→ $q = 50k-0.156k^2$

$= 50* 160.3-0.156* (160.3)^2$

$= 4006.4 \ veh/h$

(b)

The speed will be:

→ $U = (q)/(k)$

$= (4006.4)/(160.3)$

$= 25 \ mph$

(c)

The density be:

→ $1001.6 = 50k-0.156k^2$

$0.156k^2-50k+1001.6 =0$

$k = 21.5 \ veh/mi \ or \ 299 \ veh/mi$

Thus the responses above are correct.

Find out more information about density here:

brainly.com/question/6838128

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 28.0 mi/gal , how many liters of gasoline would you need to buy to complete a 142-km trip in Europe

Answers

Answer:

The volume required is $V = 11.91 \ liters$

Explanation:

From the question we are told that

The cars mileage is v = 28.0 mi/gal

The distance is d = 142 km

Converting the distance from km to miles

$d = 142 * 0.6214 = 88.24 \ miles$

Generally the volume of gasoline needed is mathematically represented as

$V = (d)/( v)$

=> $V = (88.24)/( 28.0 )$

=> $V = 3.151 \ gal$

Converting to Liters

=> $V = 3.151 * 3.78$

=> $V = 11.91 \ liters$

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of

Answers

Complete Question

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = (1)/(2)mv^2$

Making v the subject

$v = \sqrt{[(2 \Delta V * q )/(m)] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{(2 * 5.9 *10^3 * 1.60218*10^(-19) )/(9.10939 *10^(-31]) )$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $(mv^2)/(r)$ ] and this is mathematically represented as

$Bqv = (mv^2)/(r)$

making B the subject

$B = (mv)/(rq)$

r is the radius with a value = 5.4cm = $= (5.4)/(100) = 5.4*10^(-2) m$

Substituting values

$B = (9.1039 *10^(-31) * 4.556 *10^7)/(5.4*10^-2 * 1.60218*10^(-19))$

$= 0.0048 T$

In the study of sound, one version of the law of tensions is:f1= f2 √ (F1/F2)

If f1= 300, F2= 60, and f2=260, find f1 to the nearest unit.

Answers

Answer:

F1 = 80

Explanation:

f1= f2 √ (F1/F2)

Where f1 = 300, f2 = 260 and F2 = 60

Putting in the above formula

300 = 260√(F1/60)

Dividing both sides by 260

=> 1.15 = √(F1/60)

Squaring both sides

=> 1.33 = F1/60

Multiplying both sides by 60

=> F1 = 80

Please use Gauss’s law to find the electric field strength E at a distance r from the center of a sphereof radius R with volume charge density ???? = cr 3 and total charge ????. Your answer should NOT contain c. Be sure to consider regions inside and outside the sphere.

Answers

Answer:

See the explaination for the details.

Explanation:

Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

Please kindly check attachment for the step by step explaination of the answer.

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