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Hunter-Best [27]

3 years ago

11

Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's

numbers on one line, each number followed by one space. (2 pts)
Ex:Enter weight 1:236.0Enter weight 2:89.5Enter weight 3:142.0Enter weight 4:166.3Enter weight 5:93.0You entered: 236.000000 89.500000 142.000000 166.300000 93.000000

1 answer:

atroni [7]3 years ago

6 0

Answer:

See below in the explanation section the Matlab script to solve the problem.

Explanation:

prompt='enter the first weight w1: ';

w1=input(prompt);

wd1=double(w1);

prompt='enter the second weight w2: ';

w2=input(prompt);

wd2=double(w2);

prompt='enter the third weight w3: ';

w3=input(prompt);

wd3=double(w3);

prompt='enter the fourth weight w4: ';

w4=input(prompt);

wd4=double(w4);

prompt='enter the first weight w5: ';

w5=input(prompt);

wd5=double(w5);

x=[wd1 wd2 wd3 wd4 wd5]

format short

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A metal rod is 0.600 m in length at a temperature of 15.0∘C. When you raise its temperature to 37.0∘C, its length increases by 0

Katyanochek1 [597]

Answer:

The coefficient of linear expansion of the metal is ∝ = 2.91 x 10⁻⁵ °C⁻¹.

Explanation:

We know that Linear thermal expansion is represented by the following equation

Δ L = L x ∝ x Δ T ---- (1)

where Δ L is the change in length, L is for length, ∝ is the coefficient of linear expression and Δ T is the change in temperature.

Given that:

L = 0.6 m

T₁ = 15° C

T₂ = 37° C

Δ L = 0.28 mm

∝ = ?

Solution:

We know that Δ T = T₂ ₋ T₁

Putting the values of T₁ and T₂ in above equation, we get

Δ T = 37 - 15

Δ T = 22 °C

Also Δ L = 0.28 mm

Converting the mm to m

Δ L = 0.00028 m

Putting the values of Δ T, Δ L, L in equation 1, we get

0.00028 = 0.6 x ∝ x 22

Rearranging the equation, we get

∝ = 0.00028 / (0.6 x 16)

∝ = 0.00028 / 13.2

∝ = 2.12 x 10⁻⁵ °C⁻¹

4 0

4 years ago

Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh

Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

$R_a=e^{-\left ( \frac{T-r}{n}\right )B}$

$R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}$

$R_a=e^{-4.4}=0.01227$

For B

$R_b=e^{-\left ( \frac{300-336}{100}\right )3}$

$R_b=e^{1.08}=2.944$

B is better for 100 hours

(b)For 750 hours

$R_a=e^{-0.5866}=0.55621$

$R_b=e^{0.144}=1.154$

So here B is more Reliable.

3 0

3 years ago

Refrigerant R-12 is used in a Carnot refrigerator

Sphinxa [80]

Answer:

Heat transferred from the refrigerated space = 95.93 kJ/kg

Work required = 18.45 kJ/kg

Coefficient of performance = 3.61

Quality at the beginning of the heat addition cycle = 0.37

Explanation:

From figure

$Q_H$ is heat rejection process

$Q_L$ is heat transferred from the refrigerated space

$T_H$ is high temperature = 50 °C + 273 = 323 K

$T_L$ is low temperature = -20 °C + 273 = 253 K

$W_{net}$ is net work of the cycle (the difference between compressor's work and turbine's work)

Coefficient of performance of a Carnot refrigerator $(COP_{ref})$ is calculated as

$COP_{ref} = \frac{T_L}{T_H - T_L}$

$COP_{ref} = \frac{253 K}{323 K - 253 K}$

$COP_{ref} = 3.61$

From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table

$Q_H = 122.5 kJ/kg$

From coefficient of performance definition

$COP_{ref} = \frac{Q_L}{Q_H - Q_L}$

$Q_H \times COP_{ref} = (COP_{ref} + 1) \times Q_L$

$Q_L = \frac{Q_H \times COP_{ref}}{(COP_{ref} + 1)}$

$Q_L = \frac{122.5 kJ/kg \times 3.61}{(3.61 + 1)}$

$Q_L = 95.93 kJ/kg$

Energy balance gives

$W_{net} = Q_H - Q_L$

$W_{net} = 122.5 kJ/kg - 95.93 kJ/kg$

$W_{net} = 26.57 kJ/kg$

Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)

$x = \frac{s_1 - s_f}{s_g - s_f}$

From figure

$s_1 = s_4 = 1.165 kJ/(K kg)$

Replacing with table values

$x = \frac{1.165 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}{1.571 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}$

$x = 0.37$

Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get

$h_1 = h_f + x \times (h_g - h_f)$

$h_1 = 181.6 kJ/kg + 0.37 \times 162.1 kJ/kg$

$h_1 = 241.58 kJ/kg$

By energy balance, $W_{t}$ turbine's work is

$W_{t} = |h_1 - h_4|$

$W_{t} = |241.58 kJ/kg - 249.7 kJ/kg|$

$W_{t} = 8.12 kJ/kg$

Finally, $W_{c}$ compressor's work is

$W_{c} = W_{net} + W_{t}$

$W_{c} = 26.57 kJ/kg + 8.12 kJ/kg$

$W_{c} = 34.69 kJ/kg$

6 0

3 years ago

____________ are calibrated to deliver one fixed volume and are ____________ than ______________. Select one: a. Transfer pipets

LiRa [457]

Answer:

b. Transfer pipets; more accurate; measuring pipets.

Explanation:

Transfer pipets can be defined as an ungraduated or graduated disposable plastic instrument which are typically used in science laboratories to transfer or deliver liquids that are small in volume (microliters) . Also, they are generally being referred to as chemical

droppers, teat pipets, eye droppers, pasteur pipets, or droppers. It is made up of an ultra-clear long, slender tube with low density polyethylene (LDPE) and an integral molded bulb. The molded bulb allows a fixed volume of the desired liquid to be dispensed (transfered) or aspirated through compression and relaxation.

Hence, transfer pipets are calibrated to deliver one fixed volume and are more accurate than measuring pipets because of its long, slender tube which limits errors.

7 0

3 years ago

A fair cubical die is thrown four times. use the binomial probability formula to calculate the probability of at least two 3's.

Marina86 [1]

By using the binomial probability formula, the probability of at least two 3's is equal to 0.1319.

<h3>How to calculate the probability of at least two 3's?</h3>

Since the fair cubical die is thrown four times, the number of times is given by n = 4 and the probability of 3's is P = 1/6.

Mathematically, the binomial probability formula is given by:

$P(X \geq x) =\sum^{n}_{r=x} ^nC_r (p)^r (q)^{(n-r)}\\\\P(X \geq 2) = \; ^4C_2 (\frac{1}{6} )^2 (\frac{5}{6} )^{(4-2)} + ^4C_3 (\frac{1}{6} )^3 (\frac{5}{6} )^{(4-3)} + ^4C_4 (\frac{1}{6} )^4 (\frac{5}{6} )^{(4-4)}\\\\P(X \geq 2) = 6 \times \frac{1}{36} \times \frac{25}{36} + 4 \times \frac{1}{216} \times \frac{5}{36} + 1 \times \frac{1}{1296} \times 1\\\\P(X \geq 2) = \frac{150}{1296} + \frac{20}{1296}+ \frac{1}{1296}$

P(X ≥ 2) = 19/144

P(X ≥ 2) = 0.1319.

Read more on probability here: brainly.com/question/25870256

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6 0

2 years ago