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4 years ago

9

Write a loop to print all elements in hourly_temperature. Separate elements with a -> surrounded by spaces. Sample output for

the given program with input: '90 92 94 95' 90 -> 92 -> 94 -> 95 Note: 95 is followed by a space, then a newline

1 answer:

ELEN [110]4 years ago

6 0

The answer & explanation for this question is given in the attachment below.

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Given that the company "Slate Rock and Gravel" has business locations in Granitetown and Bedrock, suppose that the company has b

kotegsom [21]

Answer:

Check the explanation

Explanation:

As Granitetown requires 5000 IP address and the starting IP address is 192.172.0.0 , we have to perform subnetting using number of hosts required

Number of IP address required by Granitetown=4000

Number of hosts bits =

2n -2= 5000

n=13 (approximately)

Number of hosts bits are 13 therefore subnet prefix =(32-13)=19

We have used 32 because IP address is of 32-bit

Subnet mask corresponding to /19 = 255.255.224.0

Starting IP address (given)= 192.172.0.0

Now to calculate last IP address of the subnet, OR operation will be performed between First IP address of the subnet and !Netmask

First IP address of subnet ( in binary) =

11000000.10101100.00000000.00000000

OR

00000000.00000000.00011111.11111111 (!Netmask)

11000000.10101100.000 11111.11111111 (result)

Converting result into decimal gives IP address 192.172.31.255 which will be the broadcast IP address of the subnet therefore

Last usable IP address = 192.172.31.254

Now, bedrock requires 2000 IP address

Therefore number of host bits are =

2n -2 =2000

n=10 (approximately)

Number of hosts bits required are 10

Therefore Netmask prefix = (32 -10) =22

Now, using the next subnet for bedrock therefore the first IP address of subnet will be 192.172.32.0

Now, similarly calculating last IP address of subnet

First IP address in binary =

11000000.10101100.00100000.00000000

OR

00000000.00000000.00000011.11111111 (!Netmask)

11000000.10101100.00100011.11111111 (result)

Now, result in decimal = 192.172.35.255

Therefore ,

Broadcast address of subnet =192.172.35.255

Last usable IP address =192.172.35.254

8 0

4 years ago

A special-purpose 30-horsepower electric motor has an efficiency of 9090%. Its purchase and installation price is $2 comma 20

Taya2010 [7]

Answer:

Motor 1 recommended

Given Information:

Two 30-horsepower motors

Consumption = 4,000 hours per year

Electricity cost = $0.10 per kWh

Load factor of motors = 60%

MARR = 15%

Study period = 8 years

Motor 1:

Efficiency = 90%

Purchase + Installation cost = $2,200

Motor 2:

Efficiency = 93%

Purchase + Installation cost = $3,200

Required information:

Which motor should be recommended under these conditions = ?

Solution:

Electricity cost = Rate of electricity*Units consumed*Load factor

Electricity cost: Motor 1

Units consumed = (30 hp/0.90)*(0.746 kW/hp)*(4000 h/year)

Units consumed = 99,466.66 kWh/year

Electricity cost = (99,466.66 kWh/year)*(0.10/kWh)*(0.60)

Electricity cost = $5,968 per year

Electricity cost: Motor 2

Units consumed = (30 hp/0.93)*(0.746 kW/hp)*(4000 h/year)

Units consumed = 96,258 kWh/year

Electricity cost = (96,258 kWh/year)*(0.10/kWh)*(0.60)

Electricity cost = $5,775 per year

Present Worth Analysis: Motor 1

PW = -$2,200 - $5,968( P/A, 15%, 8)

where (P/A,15%,8) = Uniform series present worth at 15% MARR and n = 8 years

P/A = (1 + i)ⁿ - 1/ i*(1 + i)ⁿ

P/A = (1 +0.15)ⁿ - 1/ 0.15(1 + 0.15)ⁿ

P/A = 4.487

PW = -$2,200 - $5,968(4.487)

PW = -$28,978

Present Worth Analysis: Motor 2

PW = -$3,200 - $5,775( P/A, 15%, 8)

P/A = 4.487

PW = -$3,200 - $5,775(4.487)

PW = -$29,112

Recommendation:

In present worth analysis, we select the alternative which has largest PW value i.e more positive or least negative.

Since the PW of motor 1 (90% efficiency) is greater (less negative) than the motor 2 (93% efficiency).

Therefore, Motor 1 recommended

4 0

4 years ago

Evaluate each of the following and express with an appropriate prefix: (a) (430 kg)^2 (b) (0.002 mg)^2, and (c) (230 m)^3

Mekhanik [1.2K]

Answer:

a)0.185g^2

b)4μg^2( Micrograms)

c)12.167hm^3( cubic hectometer)

Explanation:

To solve this problem, we must first use the calculator to solve each of the points, then knowing the number of zeros on the left or on the right, we choose a unit with which it is better to see the unit

A)

(430 kg)^2=184900kg^2=0.185g^2

B)

(0.002 mg)^2=4x10^-6mg^2=4μg( Micrograms)

c)

(230 m)^3=12167000m^3=12.167hm^3( hectometer)

6 0

3 years ago

Exercise 5.Water flows in a vertical pipe of 0.15-m diameter at a rate of 0.2 m3/s and a pressureof 275 kPa at an elevation of 2

wariber [46]

Answer:

a. Pressure head: 33.03,

Velocity Head: 6.53

b. Pressure Head: -1.97,

Velocity Head: 6.53

Explanation:

a.

Given

Diameter = 0.15-m, radius = 0.075

rate = 0.2 m3/s

Pressure =275 kPa

elevation =25 m.

We'll consider 3 points as the water flow through the pipe

1. At the entrance

2. Inside the pipe

3. At the exit

At (1), the velocity can be found using continuity equation.

V1 = ∆V/A

Where A = Area = πr² = π(0.075)² = 0.017678571428571m²

V1 = 0.2/0.017678571428571

V1 = 11.32 m/s

The value of pressure at point 1, is given by Bernoulli equation between point 1 and 2:

P1/yH20 + V1²/2g + z1 = P2/yH20 + V2²/2g + z2

Substitute in the values

P1/yH20 + 20 = (275 * 10³Pa)/yH20 + 25

P1/yH20 = (275 * 10³Pa)/yH20 + 25 - 25

=> P1/yH20 = (275/9.81 + 5)

P1/yH20 = 33.03

The velocity head at point one is then given by

V2²/2g = 11.32²/2 * 9.8

V2²/2g = 6.53

b.

The value of pressure at point 1, is given by Bernoulli equation between point 1 and 3:

P1/yH20 + V1²/2g + z1 = P3/yH20 + V3²/2g + z3

Substitute in the values

33.03 + 20 = P3/yH20 + 55

P3/yH20 = 33.03 + 20 - 55

=> P1/yH20 = -1.97

The velocity head at point three is then given by

V2²/2g = V3²/2g = 6.53

4 0

3 years ago

Read 2 more answers

Technician A says that in a recirculating-ball steering box when the sector gear and nut teeth

NISA [10]

The second one I believe

3 0

2 years ago

Read 2 more answers