Convert the angles of a triangle to radians.Part A31∘43′53′′, 90∘32′11′′, 57∘43′56′′Express your answers, separated by commas, to six significant figures.nothingrad, rad, radRequest AnswerPart B94∘22′19′′, 40∘54′53′′, 44∘42′48′′Express your answers, separated by commas, to six significant figures.

Answers

Answer 1
Answer:

Answer:

Explanation:

To convert to radians

A31∘43′53′′, 90∘32′11′′, 57∘43′56′′

using DMS approach ; 1degree = 60minutes = 3600 seconds

1° = 60' = 3600"

And degree to radian = multiply by π/180

A) 31∘43′53′′ = 31degree + 43minutes + 53 seconds

= 31 degree + 43minutes + 53/60

= 31 degree + 43.88minutes

= 31 degree + 43.88/60 = 31.73 degree x π/180 = 0.5534radians

FOR 90∘32′11′′ = 90 degree + 32minutes + 11seconds

= 90degree + 32minutes + 11/60

= 90 degree + 32.183minutes

= 90 degree + 32.183/60 = 90.54degree x π/180

= 1.580radians

FOR 57∘43′56′′ = 57degree + 43minutes+ 56seconds

= 57degree + 43minutes + 56/60

= 57 degree + 43.93minutes

= 57degree + 43.93/60 = 57.73degree X π/180  

= 1.00radians

PART B

FOR 94∘22′19′′ = 94degree + 22minutes + 19seconds

= 94degree + 22minutes + 19/60

= 94degree + 22.32minutes

= 94degree + 22.32/60

= 94.37degree X π/180  = 1.65radians

FOR 40∘54′53′′ = 40degree + 54minutes + 53seconds

= 40 degree + 54minutes + 53/60

= 40 degree + 54.88minutes = 40 degree + 54.88/60

= 40.91degree X π/180  = 0.714radians

FOR 44∘42′48′′ = 44degree + 42minutes + 48seconds

= 44degree + 42.8minutes

= 44.71degree X π/180 = 0.780radians

Answer 2
Answer:

Answer:

A.

0.176270π rad, 0.502980π rad, 0.320735π rad

B.

0.524289π rad, 0.227304π rad, 0.248407π rad

Explanation:

We know that,

1° = 60' 180° = π

1 ' = 1°/60 1° = π/180

A.

a. 31°43'53''

Step 1

53'' = 53 * 1/60

= 53'/60

Step 2

43'53''

= 43'+53'/60

= (2580+43)/60

= 2623'/60

-------- Convert to degrees

= 2623/60 * 1/60

= 2623/3600

Step 3

31°43'53''

= 31+ 2623/3600

= (111600 + 2623)/3600

= 114223°/3600

Now, we convert to radians

= 114223/3600 * π/180°

= 0.176270π rad

b.

90°32'11''

Step 1.

11' = 11 * 1/60

= 11/60

Step 2

32'11'

= 32 + 11/60

= 1931/60

-------- Convert to degrees

= 1931/60 * 1/60

= 1931/3600

Step 3

90°31'11''

= 90 + 1931/3600

= 325931°/3600

Now we convert to radians

= 325931°/3600 * π/180°

= 0.502980π rad

c.

57°43'56''

Step 1

56' = 56 * 1/60

= 56/60

= 14/15

Step 2

43'56''

= 43 + 14/15

= 659/15

Now we convert to degrees

= 659/15 * 1/60

= 659°/900

Step 3

57°43'56''

= 57 + 659/900

= 51959/900

Now we convert to radians

= 51959°/900 * π/180°

= 0.320735π rad

B.

a.

94∘22′19′′

Step 1

19'' = 19/60

Step 2

22'19''

= 22 + 19/60

= 1339/60

Now we convert to degrees

= 1339/60 * 1/60

= 1339°/3600

Step 3

94°22'19"

= 94 + 1339/3600

= 339739°/3600

Now we convert to radians

= 339739°/3600 * π/180

= 0.524289π rad

b.

40∘54′53′′

Step 1

53" = 53/60

Step 2

54'53"

= 54'+ 53/60

= 3293/60

Now we convert to degrees

= 3293/60 * 1/60

= 3293/3600

Step 3

40°54'53"

= 40 + 3293/3600

= 147293/3600

Now we convert to radians

= 147293/3600 * π/180

= 0.227304π rad

c.

44∘42′48′

Step 1

48' = 48/69

= 4/5

Step 2

42'48"

= 42 + 4/5

=214/5

Nowz we convert to degrees

= 214/5 * 1/60

= 107/150

Step 3

44°42'48"

= 44 + 107/150

= 6707/150

Now we convert to radians

= 6707/150 * π/180

= 0.248407π rad


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1. Sewage-treatment plant, a large concrete tank initially contains 440,000 liters liquid and 10,000 kg fine suspended solids. To flush this material out of the tank, water is pumped into the vessel at a rate of 40,000 liter/h. Liquid containing solids leaves at the same rate. Estimate the concentration of suspended solids in the tank at the end of 5 h.

Answers

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

(d) Arches NP is known for its spectacular arches that develop in the jointed areas of the park. Placemark Problem 2d flies you to Landscape Arch, the arch with the largest span in Arches NP. If the stresses that stretched the rock to form the joints were oriented perpendicular to the joint surfaces and the rock fins that contain the arches, what was the direction that the rocks were stretched? ☐ N-S
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☐ NW-SE
☐ NE-SW

Answers

Answer:

☐ NE-SW

Explanation:

Based on the description, the rock direction is North East - South West (NE-SW). Rocks generally can expand or compress depending on the type and magnitude of stress applied on the rocks. However, if the applied stress is sufficiently high, cracks and fractures will be created on the rock and it can ultimately lead to the formation of particles.

What is the minimum hose diameter of an ideal vacuum cleaner that could lift a 14 kg dog off the floor?

Answers

The diameter would be 267km speed suction i think

1. An automobile travels along a straight road at 15.65 m/s through a 11.18 m/sspeed zone. A police car observed the automobile. At the instant that the two
vehicles are abreast of each other, the police car starts to pursue the automobile at
a constant acceleration of 1.96 m/s². The motorist noticed the police car in his rear
view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s². (Hint: The police will not go against the law.)
a) Find the total time required for the police car to overtake the automobile.
b) Find the total distance travelled by the police car while overtaking the
automobile.
c) Find the speed of the police car at the time it overtakes the automobile.
d) Find the speed of the automobile at the time it was overtaken by the police car.

Answers

Answer:

a.) Time = 17.13 seconds

b.) 31.88 m

c.) V = 11.18 m/s

d.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

 At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 × 1.96 ×S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.

We will use third equation of motion to solve that.

V^2 = U^2 + 2as

V^2 = 15.65^2 - 2 × 3.05 × 31.88

V^2 = 244.9225 - 194.468

V = sqrt( 50.4545)

V = 7.1 m/s

A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.

Answers

Answer:

\sigma _1=10.33MPa

\sigma _2=5.16MPa

Explanation:

Given that

Diameter(d)=62 mm

Thickness(t)= 300 μm=0.3 mm

Internal pressure(P)=100 KPa

Actually there is no any shear stress so normal stress will become principle stress.This is the case of thin cylinder.The stress in thin cylinder are hoop stress and longitudinal stress .

The hoop stress

\sigma _h=(Pd)/(2t)

Longitudinal stress

\sigma _l=(Pd)/(4t)

Now by putting the values

\sigma _h=(Pd)/(2t)

\sigma _h=(100* 62)/(2* 0.3)

\sigma _h=10.33MPa

\sigma _l=5.16MPa

So the principle stress are

\sigma _1=10.33MPa

\sigma _2=5.16MPa

When considering free energy change, biochemists usually define a standard state, the biochemical standard state, which is modified from the chemical standard state to fit biochemical applications. Determine which of the phrases describe the biochemical standard state, the chemical standard state, or both.

Answers

Answer:

Maximum work under this condition (∆G) = Maximum work under Standard Condition (∆G°) + Activities defining this condition

Explanation:

In this equation, the term DGo provides us with a value for the maximal work we could obtain from the reaction starting with all reactants and products in their standard states, and going to equilibrium. The term DG' provides us with a value for the maximal work we could obtain under the conditions defined by the activities in the logarithmic term. The logarithmic term can be seen as modifying the value under standard conditions to account for the actual conditions. In describing the work available in metabolic processes, we are concerned with the actual conditions in the reaction medium (whether that is a test-tube, or the cell cytoplasm); the important term is therefore DG'. If we measure the actual activities (in practice, we make do with concentrations), and look up a value for DGo in a reference book, we can calculate DG' from the above equation.

Values for DGo provide a useful indication through which we can compare the relative work potential from different processes, because they refer to a standard set of conditions.

Therefore both phrases describe the Biochemical and Chemical Standard State