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Ainat [17]
3 years ago
13

If you're carrying a load that extends beyond the sides of your vehicle, you don't need to clearly mark it unless it extends mor

e than _____ beyond the rear of the vehicle.
A. four inches
B. four feet
C. six feet
D. eight feet
Engineering
2 answers:
OlgaM077 [116]3 years ago
5 0

Answer:

B. four feet

Explanation:

If you're carrying a load that extends beyond the sides of your vehicle, you don't need to clearly mark it unless it extends more than four feet beyond the rear of the vehicle.

Serjik [45]3 years ago
3 0

Answer:

Your corrct answer is B. four feet

Explanation:

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I need unseen passage
monitta

Answer:

this unseen passage__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

8 0
3 years ago
Read 2 more answers
Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor i
Temka [501]

Answer:

Explanation:

Find the temperature at exit of compressor

T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k

Find the work done by the compressor

\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg

Find the actual workdone by the compressor

\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg

Find the temperature at exit of the turbine

T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k

Find the actual workdone by the turbine

1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg

Find the temperature of the regeneration

\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k

Find the heat supplied

Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg

Find the thermal efficiency

n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4

60.4%

Find the mass flow rate

m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42

Find the actual workdone by the compressor

\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg

Find the actual workdone by the turbine

\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg

Find the temperature of the compressor exit

\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k

Find the temperature at the turbine exit

4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k

Find the temperature of regeneration

\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k

6 0
3 years ago
Read 2 more answers
What is the impedance of a 5μF capacitor at a frequency of 500Hz? What is the impedance of a60mH inductor at this frequency?
choli [55]

Answer:

1) 63.66 ohm

2) 188.49 ohm

Explanation:

Data provided in the question:

Part 1

Capacitance, C = 5μF = 5 × 10⁻⁶ F

Frequency = 500 Hz

Now,

Impedance = \frac{1}{2\times\pi\times f\times C}

or

Impedance = \frac{1}{2\times\pi\times500\times5\times10^{-6}}

or

Impedance = 63.66 ohm

Part 2

Inductance = 60 mH = 60 × 10⁻³ H

Frequency = 500 Hz

Now,

Impedance for an inductor = 2πfL

thus,

Impedance = 2 × π × 500 × 60 × 10⁻³

= 188.49 ohm

4 0
3 years ago
The Accenture team is involved in helping a client in the transformation journey using Cloud computing. How is myNav beneficial
professor190 [17]

Answer:

It helps ensure the security of client data.

Explanation:

Cloud computing is one of the security features in networking which is being adopted by big corporations and organizations inorder to ensure smooth running of the organization.

Also, due to the sensitivity of data which some organizations deals in, they tried everything possible to protect themselves and their clients' information through use of cloud computing.<em> Data are stored in the clouds, and requires special administrative rights for it to be accessed by some of the staffs working in any given organization.</em>

8 0
3 years ago
A plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service facto
vazorg [7]

Answer:

(M_t)_{rated}=61.11lb-in

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = 4 \times 0.7457 =2.9828 KW

service factor(k)= 2.75

now,

KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }

M_t= 18,989.09 \ N-mm= 168.06 lb-in

torque rating

(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in

4 0
4 years ago
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