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ValentinkaMS [17]
3 years ago
15

CAN SOMEONE GIVE ME AND ANSWER AND EXPLANTION FOR ALL THESE QUESTIONS PLEASE, I AM STRUGGLING

Engineering
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer:

13.) 44V

14.) 66V

15.) 9A

16.) 12V

Explanation:

Concepts you need to know:

1. Resistors in series can be combined by summing up their values.

2. The current always stays the same when the resistors are in series.

3. Ohms Law:   V = IR

13.) We know that the resistors are in series so we can combine them into a single big resistor. We got R₄ = R₃+R₂+R₁ = 2+4+5 = 11Ω.

We also know that current is the same in a series of resistors, so there's 4A going through the entire circuit.

Using ohm's law, we can calculate V = IR = (4)(11) = 44V

14.) Same explanation as part (13). Except now the current is I=6A. All we have to do is to replace the previous I=4 to I=6.

V = IR = (6)(11) = 66V

15.) Again, we know that current is the same when resistors are in series. So if you can find the current through R1 then that's the same current throughout the entire circuit.

We're given E₁=18V, so...

I = V/R = E₁/R₁ = 18/2 = 9A

this means 9A will also flow through R₃.

16.) Same concept as part (15). We know that E₃=15V, so lets find the current.

I = V/R = E₃/R₃ = 15/5 = 3A

Since current is same throughout the circuit, I₁=3A as well. Use this to calculate the voltage through R₁.

V = IR = I₁R₁ = (3)(2) = 6V

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5 0
3 years ago
The 40-ft-long A-36 steel rails on a train track are laid with a small gap between them to allow for thermal expansion. Determin
vfiekz [6]

Answer:

Ф = 0.02838 ft

F  = 1,032 N

Explanation:

To find out gap delta,

As it is case of free thermal expansion,

First we start with, some assumptions we have to made to solve this problem.

1. Thermal Expansion Coefficient of Steel is ∝= 6.45 ×10^(-6)

2. Modulas of elasticity for A-36 steel is E= 200 GPa

3. Area of rail is assumed to be unit area.

The gape required can be given by,

Ф = ∝  × ΔT  × L  ... where Ф= Gap Delta in ft

                                          ΔT= Temperature rise in F

                                               = 90- (-20)

                                               =  110 F

Ф =  6.45 ×10^(-6) × 110 × 40

Ф =  28,380 × 10^(-6) ft

Ф = 0.02838 ft     .... total gape required for expansion of steel rails

Stress induced in rails is given by,

   σ     =  ∝  × ΔT  × E

          =  6.45 ×10^(-6)   × 110  × 200

  σ      =  1,41,900 Pa

Now, let's find axial force in rails,

Here,we have to consider  ΔT= 20 F.

As due to temperature change, axial force generated in rails can be find by,

F = A × ∝ × ΔT× E × L

F = 1 × 6.45 × 10^(-6) × 20 × 200 × 10^(-9) × 40

F = 25,800 × 40 × 10^(-3)

F = 10,32,000 × 10^(-3)

F= 1,032 N

Finally, due to temperature change, rail is subjected to axial force, axial stress.

8 0
3 years ago
What does this work for
Anastaziya [24]

Answer:

it allows your dash board to light up you MPH RPM and all the other numbers on the spadomter

Explanat

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3 years ago
Teresa is emotionally volatile, particularly with friends and boyfriends. She is extremely dramatic about even the smallest disa
Ilia_Sergeevich [38]

Answer: Borderline personality Disorder.

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Its symptoms includes unstable emotions, sense of insecurity, worthlessness, and impulsivity.

This condition can not be cured, but treatments such as therapies, medication (in some cases) could help.

4 0
3 years ago
Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large
salantis [7]

Answer:

Q_{cv} = -1007.86kJ

Explanation:

Our values are,

State 1

V=3m^3\\P_1=1bar\\T_1 = 295K

We know moreover for the tables A-15 that

u_1 = 210.49kJ/kg\\h_i = 295.17kJkg

State 2

P_2 =6bar\\T_2 = 296K\\T_f = 320K

For tables we know at T=320K

u_2 = 228.42kJ/kg

We need to use the ideal gas equation to estimate the mass, so

m_1 = \frac{p_1V}{RT_1}

m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}

m_1 = 3.54kg

Using now for the final mass:

m_2 = \frac{p_2V}{RT_2}

m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}

m_2 = 19.59kg

We only need to apply a energy balance equation:

Q_{cv}+m_ih_i = m_2u_2-m_1u_1

Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i

Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)

Q_{cv} = -1007.86kJ

The negative value indidicates heat ransfer from the system

7 0
3 years ago
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