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Nastasia [14]
3 years ago
12

9 Select the correct answer. Conservation agents object to legal hunting. OA True B. False​

Engineering
2 answers:
Aleksandr-060686 [28]3 years ago
8 0
False

Conservation agents are dependent on legal hunting. License fees pay their salaries, fund conservation projects etc.
m_a_m_a [10]3 years ago
5 0

Answer:

its true

Explanation:

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3 0
3 years ago
Question 9 Two cells are connected in parallel. All cells are rated 1.5 V at 10 A. Et = ? (V)​
Eddi Din [679]
Parallel hookups increase amp hour capacity but voltage remains the same... 1.5 volts

Series the two and voltage is 3.0 volts.
7 0
4 years ago
Stainless steel ball bearings (rho = 8085 kg/m3 and cp = 0.480 kJ/kg·°C) having a diameter of 1.2 cm are to be quenched in water
sveticcg [70]

Answer:

\dot Q = -2.341\,kJ

Explanation:

The rate of heat transfer from the balls to the air is:

\dot Q = \left[(800\,\frac{1}{min} )\cdot (\frac{1\,min}{60\,s} )\cdot(8085\,\frac{kg}{m^{3}})\cdot (\frac{4}{3}\pi )\cdot (6\times 10^{-3}\,m)^{3}\right]\cdot (0.480\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (850\,^{\textdegree}C-900\,^{\textdegree}C)\dot Q = -2.341\,kJ

3 0
4 years ago
Read 2 more answers
The bolts that attach a bracket to an industrial machine must each carry a static tensile load of 4 kN.
HACTEHA [7]

Answer:

M10 × 1.5

least number of threads is 4.3

Explanation:

given data

static tensile load = 4 kN = 4000 N

safety factor = 5

coarse‐thread metric = 5.8

solution

we know that proof load for 5.8  class is Sp = 380 MPa

so area of cross section is express as

area = \frac{force \times FOS}{Sp}    ......................1

area = \frac{4000 \times }{380}

area = 52.6  mm²

so by table at 58 mm² we can say we resist  M10 × 1.5

and

bolt tensile strength is express as

bolt tensile strength = nut shear strength    ......2

so here bolt tensile strength = At × Sy (bolt)    ...............3

and nut shear strength =  πd ( 0.75 t ) Sys

nut shear strength =   π × 10 × ( 0.75 t )  × 0.58  × \frac{2}{3}   × Sy(bolt)   ............4

so from equation 3 and 4 we get

t = 6.37 mm

and

for the pitch is = 1.5 mm

least number of threads is 4.3

8 0
3 years ago
Fluid originally flows through a tube at a rate of 100 cm^3/s. To illustrate the sensitivity of the Poiseuille flow rate to vari
ankoles [38]

Answer:

a) 0.01

b) 150 cm^3/s

c) 300 cm^3/s

d) 25 cm^3/s

Explanation:

a) We know that :  

     Q=ΔP/R

     R=8ηl/π*r^4

Givens:

     r^2 = 0.1 r_1  

Plugging known information to get :  

      Q=ΔP/R

         =ΔP*π*r^4/8*η*l

Q_2/r_2^4 =Q_1/r_1^4

           Q_2=Q_1/r_1^4*r_2^4

                  =Q_1/r_1^4*r*0.0001*r_1^4

          Q_2 = 0.01

b) From the rate flow of the fluid we know that :  

            Q=ΔP/R                                   (1)

              F=η*Av/l                                 (2)

             R=8*ηl/π*r^4                           (3)

<em>Where: </em>

ΔP is the change in the pressure .  

r is the raduis of the tube .

l is the length of the tube .

η is the coefficient of the vescosity of the fluid .

R is the resistance of the fluid .

Givens: Q1 = 100 cm^3/s , ΔP= 1.5

Plugging known information into EQ.1 :  

                 Q=ΔP/R

     Q_2/ΔP2=Q_1/ΔP

              Q_2=150 cm^3/s

c) we know that :

      F = η*Av/l  

can be written as :  

     ΔP = F/A = η*v/l  

Givens: η_2 = 3η_1  

           Q=ΔP/R  

           Q=η*v/l*R

Q_2/η_2=Q_1/η_1

       Q_2=300 cm^3/s

d) We know that :  

           Q=ΔP/R

           R=8*ηl/π*r^4  

Givens: l_2 = 4*l_1

Plugging known information to get :  

              Q=ΔP/R  

              Q=ΔP*π*r^4/8*ηl

   Q_2/l_2=Q_1/l_1

          Q_2 = 25 cm^3/s

5 0
4 years ago
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