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4 years ago

Ame: 7. A step-down transformer reduces the primary current. True or false

Engineering

2 answers:

RSB [31]4 years ago

6 0

False

..............

It increases the primary current a step up reduces the primary current

ivanzaharov [21]4 years ago

4 0

The answer is false.

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A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu

bazaltina [42]

Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

b. Area of Aluminum Foil needed = 0.45m²

c. To keep a 1.0-nF, a larger area of Teflon is required.

Explanation:

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

Where

K = 3

ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²

A = Area = 22 * 28 = 616cm² = 0.0616m²

C = 1.0-nF = 1 * 10^-12F

So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)

d = 1.64 * 10^-3m

d = 1.64mm

Now, that the distance has been solved.

The Number of Sheets, N is given by

N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm

N = 1.64/0.2

N = 8.2

N = 8 sheets --- Approximated

Here, she's changed the diameter of the sheets to 12mm

Well make use of the formula in (a) above

Using d = Kε0A/C

Where

d = 12 * 10^-3m

Other constraints remain unchanged

Make A the subject of formula

A = dC/Kε0

A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)

A= 0.45m²

c. From (b) above

A ∝ 1/K

As the dielectric constant increase, the area decreases

The dielectric constant of a Teflon is 2.1

This means that if she used a Teflon instead, the area will be larger.

So, to keep a 1.0-nF, a larger area of Teflon is required.

7 0

4 years ago

Knowing that v = –8 m/s when t = 0 and v = 8 m/s when t = 2 s, determine the constant k. (Round the final answer to the nearest

docker41 [41]

Answer:

a)We know that acceleration a=dv/dt

So dv/dt=kt^2

dv=kt^2dt

Integrating we get

v(t)=kt^3/3+C

Puttin t=0

-8=C

Putting t=2

8=8k/3-8

k=48/8

k=6

5 0

3 years ago

What do electrons have to do with electrical current?

erastova [34]

Answer:

Electrons in atoms can act as our charge carrier, because every electron carries a negative charge. If we can free an electron from an atom and force it to move, we can create electricity.

6 0

4 years ago

The assembly consists of a brass shell (1) fully bonded to a ceramic core (2). The brass shell [E = 93 GPa, α= 15.1 × 10−6/°C] h

marshall27 [118]

Answer:

ΔT = 62.11°C

Explanation:

Given:

- Brass Shell:

Inner Diameter d_i = 32 mm

Outer Diameter d_o = 39 mm

E_b = 93 GPa

α_b = 15.1*10^-6 / °C

- Ceramic Core:

Outer Diameter d_o = 32 mm

E_c = 310 GPa

α_c = 3.2*10^-6 / °C

- Unstressed @ T = 8°C

- Total Length of the cylinder L = 160 mm

Find:

Determine the largest temperature increase Δ⁢t that is acceptable for the assembly if the normal stress in the longitudinal direction of the brass shell must not exceed 60 MPa.

Solution:

- Since, α_b > α_c the brass shell is in compression and ceramic core is in tension. The stress in shell is given as б_a:

б_b = - 60 MPa

- The force equilibrium can be written as:

б_b*A_b + б_c*A_c = 0

Where, б_b is the stress in core

A_b is the cross sectional area of the shell

A_c is the cross sectional area of the core

б_b*pi*( d_o^2 - d_i^2) / 4 + б_c*pi*( d_i^2) / 4 = 0

б_b*( d_o^2 - d_i^2) + б_c*( d_i^2) = 0

б_c = - б_b*( d_o^2 - d_i^2) / ( d_i^2)

Plug in the values:

б_c = 60*( 0.039^2 - 0.032^2) / ( 0.032^2)

б_c = 29.121 MPa , б_b = - 60 MPa

- The total strains in both brass shell and ceramic core is given by:

ξ_b = α_b*ΔT + б_b / E_b

ξ_c = α_c*ΔT + б_c / E_c

- The compatibility relation is:

ξ_b = ξ_c

α_b*ΔT + б_b / E_b = α_c*ΔT + б_c / E_c

ΔT*(α_b - α_c ) = б_c / E_c - б_b / E_b

ΔT = [ б_c / E_c - б_b / E_b ] / (α_b - α_c )

Plug in values and solve:

ΔT = [ 0.029121 / 310 + 0.06 / 93 ]*10^6 / (15.1 - 3.2 )

ΔT = 62.11°C

8 0

3 years ago

Plot the data as impact energy versus temperature: (b) Determine a ductile-to-brittle transition temperature as the temperature

Shkiper50 [21]

Question:

The following tabulated data were gathered from a series of Charpy impact tests on a commercial low-carbon steel alloy.

Temperature(∘C)50403020100-10-20-30-40Impact energy (J)76767158382314951.5

(a) Plot the data as impact energy versus temperature.

(b) Determine a ductile-to-brittle transition temperature as the temperature corresponding to the average of the maximum and minimum impact energies.

(c) Determine a ductile-to-brittle transition temperature as the temperature at which the impact energy is 20 J.

Answer:

a) see the attached graph

b) max E = 76 and Min E =1.53 average = 77.5/2 = 38.75 J

this corresponds to about 10° C

c ) at E = 20 J temperature is about -2°C

8 0

4 years ago

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