1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
katen-ka-za [31]
2 years ago
8

How do you get your drivers lisnes when your 15

Engineering
1 answer:
prohojiy [21]2 years ago
5 0

Answer:

You take a drivers test!

Explanation:

You might be interested in
As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial tempera
saveliy_v [14]

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length  = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod \overline T = 548 \ K

\rho = 7900 \ kg/m^3

K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K

\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\  B_i = \dfrac{h(\rho/4)}{K} \\ \\  =0.657

Here, we can't apply the lumped capacitance method, since Bi > 0.1

\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\

0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\  0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o   \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81

t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\  t_f= 2162.5 \\ \\ t_f = 36 mins

However, on a single rod, the energy extracted is:

\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) )  \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\  \theta = 1.54 \times 10^7 \ J

Hence, for centerline temperature at 50 °C;

The surface temperature is:

T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}

5 0
3 years ago
A Sampling of 409 motor Shafts from a very large production Batch shows a sample standard deviation in Diameter of 0.021 mm with
andrezito [222]

Answer:

9.248 < \mu < 9.253 mm

Explanation:

Given data:

standard deviation  = 0.021 mm

sample mean = 9.251 mm

total sample = 409 m

confidence interval level = 95%

mean diameter of entire batch  \mu  = \bar X \pm z \frac{\sigma}{\sqrt{n}}

where

\bar X - sample mean = 9.251 mm

z = critical value

plugginf all value in the above relation to get the mean diameter

\mu = 9.251 \pm 1.96 \times \frac{0.021}{\sqrt{409}}

9.248 < \mu < 9.253 mm

7 0
3 years ago
1. Use the charges to create an electric dipole with a horizontal axis by placing a positive and a negative charge (equal in mag
White raven [17]

Answer:

2)

a)  to the right of the dipole    E_total = kq [1 / (r + a)² - 1 / r²]

b)To the left of the dipole      E_total = - k q [1 / r² - 1 / (r + a)²]

c) at a point between the dipole, that is -a <x <a  

      E_total = kq [1 / x² + 1 / (2a-x)²]

d) on the vertical line at the midpoint of the dipole (x = 0)

E_toal = 2 kq 1 / (a ​​+ y)² cos θ

Explanation:

2) they ask us for the electric field in different positions between the dipole and a point of interest. Using the principle of superposition.

This principle states that we can analyze the field created by each charge separately and add its value and this will be the field at that point

Let's analyze each point separately.

The test charge is a positive charge and in the reference frame it is at the midpoint between the two charges.

a) to the right of the dipole

The electric charge creates an outgoing field, to the right, but as it is further away the field is of less intensity

           E₊ = k q / (r + a)²

where 2a is the distance between the charges of the dipole and the field is to the right

the negative charge creates an incoming field of magnitude

           E₋ = -k q / r²

The field is to the left

therefore the total field is the sum of these two fields

           E_total = E₊ + E₋

           E_total = kq [1 / (r + a)² - 1 / r²]

we can see that the field to the right of the dipole is incoming and of magnitude more similar to the field of the negative charge as the distance increases.

b) To the left of the dipole

The result is similar to the previous one by the opposite sign, since the closest charge is the positive one

E₊ is to the left and E₋ is to the right

          E_total = - k q [1 / r² - 1 / (r + a)²]

We see that this field is also directed to the left

c) at a point between the dipole, that is -a <x <a

In this case the E₊ field points to the right and the E₋ field points to the right

                      E₊ = k q 1 / x²

                      E₋ = k q 1 / (2a-x)²

                      E_total = kq [1 / x² + 1 / (2a-x)²]

in this case the field points to the right

d) on the vertical line at the midpoint of the dipole (x = 0)

    In this case the E₊ field points in the direction of the positive charge and the test charge

    in E₋ field the ni is between the test charge and the negative charge,

the resultant of a horizontal field in zirconium on the x axis (where the negative charge is)

                      E₊ = kq 1 / (a ​​+ y) 2

                      E₋ = kp 1 / (a ​​+ y) 2

                      E_total = E₊ₓ + E_{-x}

                      E_toal = 2 kq 1 / (a ​​+ y)² cos θ

e) same as the previous part, but on the negative side

                        E_toal = 2 kq 1 / (a ​​+ y)² cos θ

When analyzing the previous answer there is no point where the field is zero

The different configurations are outlined in the attached

3) We are asked to repeat part 2 changing the negative charge for a positive one, so in this case the two charges are positive

a) to the right

in this case the two field goes to the right

           E_total = kq [1 / (r + a)² + 1 / r²]

b) to the left

            E_total = - kq [1 / (r + a)² + 1 / r²]

c) between the two charges

E₊ goes to the right

E₋ goes to the left

            E_total = kq [1 / x² - 1 / (2a-x)²]

d) between vertical line at x = 0

             

E₊ salient between test charge and positive charge

           E_total = 2 kq 1 / (a ​​+ y)² sin θ

In this configuration at the point between the two charges the field is zero

8 0
3 years ago
I am making composites of silicone rubber and copper particles; by mixing thermally conductive particles into the thermally insu
Gelneren [198K]

Answer:

The composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.

Explanation:

Let us assume that the total mass of composite is 100 lbm So as per the given conditions

  • 15 lbm is copper and 85 lbm is rubber.
  • Density of rubber is 70 lbm/ft3
  • Specific gravity of Copper is 9

So As per the formula of specific gravity

                                         S_{cu}=\frac{\rho_{cu}}{\rho_w}

here density of water is 62.4 lbm/ft3

Solving for Density of Copper gives

                                        S_{cu}=\frac{\rho_{cu}}{\rho_w}\\9=\frac{\rho_{cu}}{62.4}\\\rho_{cu}=9 \times 62.4\\\rho_{cu}=561.5 lbm/ft3

For composition on volume basis, volume of individual components and composite are calculated as

                                          V_{cu}=\frac{m_{cu}}{\rho_{cu}}\\V_{cu}=\frac{15}{561.5}\\V_{cu}=0.0267 ft^3\\\\V_{r}=\frac{m_{r}}{\rho_{r}}\\V_{r}=\frac{85}{70}\\V_{r}=1.214 ft^3\\\\V_{c}=V_{r}+V_{cu}\\V_{c}=1.214+0.0267 \\V_{c}=1.2409 ft^3

The composition is given as

c_{cu}=\frac{V_{cu}}{V_{c}}\\c_{cu}=\frac{0.0267}{1.2409} \times 100 \%\\c_{cu}=2.154 \%\\\\c_{r}=\frac{V_{r}}{V_{c}}\\c_{r}=\frac{1.214}{1.2409} \times 100 \%\\c_{r}=97.83 \%

So the composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.

6 0
3 years ago
Which is the most common source of abnormal noises in an air
inn [45]

<em>Answ</em><em>er</em><em> </em><em>is</em><em>:</em><em> </em><em>letter</em><em> </em><em>B.Condenser</em>

Explanation:

<em>common issue associated with an air conditioner screeching noise is a malfunctioning fan motor in the outdoor condenser unit. The central air conditioning system connected to your home features a fan that is designed to remove heat from the refrigerant.</em>

5 0
3 years ago
Other questions:
  • Ayuda porfavor es para una tarea de mi capacitación de desarrollo microempresarial
    14·1 answer
  • During the collision, is the magnitude of the force of asteroid A on asteroid B greater than, less than, or equal to the magnitu
    11·2 answers
  • an adiabatic compressor receives 1.5 meter cube per second of air at 30 degrees celsius and 101 kpa. The discharge pressure is 5
    11·1 answer
  • A Gaussian random voltage X volts is input to a half-wave rectifier and the output voltage is Y = Xu (X) Volts were u (x) is the
    9·1 answer
  • Please write the command(s) you should use to achieve the following tasks in GDB. 1. Show the value of variable "test" in hex fo
    5·1 answer
  • For the same cross-sectional area, which column provides the higher buckling load: a circular bar or a circular tube?
    15·1 answer
  • What are common names assigned to instruction addresses in a PLC program called
    8·1 answer
  • 1)
    13·1 answer
  • A fine-grained soil has a liquid limit of 200%, determined from the Casagrande cup method. The plastic limit was measured by rol
    15·1 answer
  • Technician A says that if fuel pump pressure is correct, fuel pump volume will be correct as well. Technician B says that a fuel
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!