Which of the following expressions causes an implicit conversion between types? Assume variable x is an integer, t is a float, and name is a string.Group of answer choices7.5 + (x / 2)x + 2 * x"Hello, " + str(name)print(str(t))

Answers

Answer 1

Answer:

x + 2 * x is the correct option. The above-selected option demonstrates implicit conversion, which is an automated type of conversion. Thus, option B is correct.

The series of conversions are necessary to change the type of a function call's argument to that of the parameter with the same name in the function declaration is known as an implicit conversion sequence. For each parameter, the compiler tries to identify an implicit conversion sequence.

If both user-defined conversion sequences A and B contain the same user-defined conversion function or constructor, and if the second standard conversion sequence of A is superior to the second standard conversion sequence of B, then user-defined conversion sequence A is preferable to user-defined conversion sequence B.

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A tensile test is carried out on a bar of mild steel of diameter 20 mm. The bar yields under a load of 80 kN. It reaches a maximum load of 150 kN, and breaks finally at a load of 70 kN. Find (i) the tensile stress at the yield point (1i) the ultimate tensile stress; (iii) the average stress at the breaking point, if the diameter of the fractured neck is 10mm

How much computer memory (in bytes) in minimum would be required to store 10 seconds of a sensor signal sampled by a 12-bit A/D converter operating at a sampling rate of 5 kHz?

Answers

Answer:

73.24 K byte

Explanation:

Assuming that

N = total number of samples

N = 10 * 5kHz

N = 50*10^3

Also, the total number of bits, T

T = 12 * N

T = 12 * 50*10^3

T = 600 * 10^3

And then, finally, the total number of byte,

B = 600*10^(3/8)

B = 75*10^3 byte

75*10^3 byte = 75*10^3/1024 kilo byte

And on converting to decimal, we will have

= 73.24 K byte

Therefore, the memory required = 73.24 K byte

Given the latent heat of fusion (melting) and the latent heat of vaporisation for water are Δhs = 333.2 kJ/kg and Δhv = 2257 kJ/kg, respectively. Use these values to estimate the total energy required to melt 100 kg of ice at 0 °C and boil off 40 kg of water at 100 °C. a) 239,028 kJ b) 95,250 kJ c) 185,500 kJ d) 362,628 kJ e) 123,600 kJ

Answers

Answer:

C)185,500 KJ

Explanation:

Given that

Latent heat fusion = 333.23 KJ/kg

Latent heat vaporisation = 333.23 KJ/kg

Mass of ice = 100 kg

Mass of water = 40 kg

Mass of vapor=60 kg

Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of vaporisation .

Sensible heat for water Q

$Q=mC_p\Delta T$

For water

$C_p=4.178\ KJ/Kg.K$

Q=4.178 x 40 x 100 KJ

Q=16,712 KJ

So total heat

Total heat =100 x 333.23+16,712 + 60 x 2257 KJ

Total heat =185,455 KJ

Approx Total heat = 185,500 KJ

So the answer C is correct.

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating

Answers

Answer:

The answer is below

Explanation:

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.

Given that:

number of poles (p) = 4, frequency (f) = 60 Hz

1) The synchronous speed of the motor is the speed at which the magnetic field of the stator is rotating. It is given as:

$n_s=(120f)/(p)=(120*60)/(4)=1800\ rpm$

2) The slip (s) = 0.05

The speed of the motor (n) is the speed of the rotor, it is given as:

$n=n_s-sn_s\n\nn=1800-0.05(1800)\n\nn=1800-90\n\nn=1710\ rpm$

3) s = 0.04

The rotor frequency is the product of the supply frequency and slip it is given as:

$f_r=sf\n\nf_r=0.04*60\n\nf_r=2.4\ Hz$

4) At standstill, the motor speed is zero hence the slip = 1:

$s=(n_s-n)/(n_s)\n \nn=0\n\ns=(n_s-0)/(n_s)\n\ns=1$

The rotor frequency is the product of the supply frequency and slip it is given as:

$f_r=sf\n\nf_r=1*60\n\nf_r=60\ Hz$

The barrel of a bicycle tire pump becomes quite warm during use. Explain the mechanisms responsible for the temperature increase.

Answers

Answer:

The air heats up when being compressed and transefers heat to the barrel.

Explanation:

When a gas is compressed it raises in temperature. Assuming that the compression happens fast and is done before a significant amount of heat can be transferred to the barrel, we could say it is an adiabatic compression. This isn't exactly true, it is an approximation.

In an adiabatic transformation:

$P^(1-k) * T^k = constant$

For air k = 1.4

$P0^(-0.4) * T0^(1.4) = P1^(-0.4) * T1^(1.4)$

$T1^(1.4) = (P1^(0.4) * T0^(1.4))/(P0^(0.4))$

$T1^(1.4) = (P1)/(P0)^(0.4) * T0^(1.4)$

$T1 = T0 * (P1)/(P0)^(0.4/1.4)$

$T1 = T0 * (P1)/(P0)^(0.28)$

SInce it is compressing, the fraction P1/P0 will always be greater than one, and raised to a positive fraction it will always yield a number greater than one, so the final temperature will be greater than the initial temperature.

After it was compressed the hot air will exchange heat with the barrel heating it up.

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam material has a specific gravity, SG, of 3.1. You may assume that the dam is loosely attached to the ground at its base, though there is significant friction to keep it from sliding.Is the weight of the dam sufficient to prevent it from tipping around its lower right corner?