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Pepsi [2]
2 years ago
15

A ball bearing has been selected with the bore size specified in the catolog as 35.000 mm to 35.020 mm. Specify appropriate mini

mum and maximum shaft diameters to provide a locational interference fit
Engineering
1 answer:
Fofino [41]2 years ago
8 0

Answer:

the minimum shaft diameter is 35.026 mm

the maximum shaft diameter is 35.042mm

Explanation:

Given data;

D-maximum = 35.020mm and d-minimum = 35.000mm

we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6

so From table, Selection of International Trade Grades metric series

the grade tolerance are;

ΔD = IT7(0.025 mm)

Δd = IT6(0.016 mm)

Also from Table "Fundamental Deviations for Shafts" metric series

Sf = 0.026

so  

D-maximum

Dmax = d + Sf + Δd

we substitute

Dmax = 35 + 0.026 + 0.016

Dmax = 35.042 mm

therefore the maximum diameter of shaft is 35.042mm

d-minimum

Dmin = d + Sf

Dmin = 35 + 0.026

Dmin = 35.026 mm

therefore the minimum diameter of shaft is 35.026 mm

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A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to
Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 =  (16/3) * 10^2 V

c)  E = 64/15 J

d)  dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the final potential difference across each capacitor

(c) the final energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

                           Q_initial = C_1*V

                           Q_initial = 20*10^-6 * 800

                           Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

                          Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

                          V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

                          Q_2 = Q_1*C_2/C_1

                          Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

                          Q_initial = 1.5*Q_1

                          Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

                          V_2 = V_1 = Q_1 / C_1  

                                            = 10.67*10^-3 / 20*10^-6

                                            = (16/3) * 10^2 V

c)

- The energy in the system is:

                          E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

                           E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

                          E = 64/15 J

d)

- The decrease in energy of the capacitors is:

                           dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

                          dE = 0.5*10^-6*20*800^2 - (64/15)

                          dE = 32/5 - 64/15 = 32/15 J

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