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3 years ago

What happens when the arms of the milky move away from the center of the galaxy

1 answer:

6 0

Well this question is though because we have never seen such a thing ! and to be quite frank when that happens , nothing good comes from it. Black Holes

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Write a program to calculate overtime pay of 10 employees. Overtime is paid at the rate of Rs. 12.00

fgiga [73]

Answer:

Here is the code.

Explanation:

#include<stdio.h>

int main()

{

int i, time_worked, over_time, overtime_pay = 0;

for (i = 1; i <= 10; i++)

{

printf("\nEnter the time employee worked in hr ");

scanf("%d", &time_worked);

if (time_worked>40)

{

over_time = time_worked - 40;

overtime_pay = overtime_pay + (12 * over_time);

}

printf("\nTotal Overtime Pay Of 10 Employees Is %d", overtime_pay);

return 0;

}

Output :

Enter the time employee worked in hr 42

Enter the time employee worked in hr 45

Enter the time employee worked in hr 42

Enter the time employee worked in hr 41

Enter the time employee worked in hr 50

Enter the time employee worked in hr 51

Enter the time employee worked in hr 52

Enter the time employee worked in hr 53

Enter the time employee worked in hr 54

Enter the time employee worked in hr 55

Total Overtime Pay Of 10 Employees Is 1020.

6 0

3 years ago

Which line from "On Becoming an Inventor" supports the idea that Dean's time at Worcester Polytechnic Institute was very useful

Stolb23 [73]

Answer:

Explanation:

The line from "On becoming an inventor" that says:

"I found that at college I could get help from my teachers with solving business problems and in learning new techniques for designing new things"

On Becoming an Inventor was by Dean Kamen an American Engineer, Inventor and Businessman

3 0

3 years ago

Read 2 more answers

A man weighs 145 lb on earth.Part ASpecify his mass in slugs.Express your answer to three significant figures and include the ap

adell [148]

Answer:

a) 4.51 lbf-s^2/ft

b) 65.8 kg

c) 645 N

d) 23.8 lb

e) 65.8 kg

Explanation:

Weight of the man on Earth = 145 lb

a) Mass in slug is...

32.174 pound = 1 slug

145 pound = $x$ slug

$x$ = 145/32.174 = 4.51 lbf-s^2/ft

b) Mass in kg is...

2.205 pounds = 1 kg

145 pounds = $x$ kg

$x$ = 145/2.205 = 65.8 kg

c) Weight in Newton = mg

where

m is mass in kg

g is acceleration due to gravity on Earth = 9.81 m/s^2

Weight in Newton = 65.8 x 9.81 = 645 N

d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,

1 m/s^2 = 3.2808 ft/s^2

$x$ m/s^2 = 5.30 ft/s^2

$x$ = 5.30/3.2808 = 1.6155 m/s^2

weight in Newton = mg = 65.8 x 1.6155 = 106

weight in pounds = 106/4.448 = 23.8 lb

e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth

mass on the moon = 65.8 kg

3 0

3 years ago

These are the most widely used tools and most often abuse tool

Mars2501 [29]

Where is the picture in this problem? How am I supposed to answer if I can’t see any footage taken from this problem.

4 0

3 years ago

Read 2 more answers

The Ethernet (CSMA/CD) alternates between contention intervals and successful transmissions. Assume a 100 Mbps Ethernet over 1 k

Vesnalui [34]

<h3>CSMA/CD Protocol: </h3>

Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.

But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.

There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.

Example:

$T_{P}=1 H r$ , at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.

12:00 AM A will see collisions

Pocket Size to detect the collision.

$\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}$

CSMA/CD is widely used in Ethernet.

Efficiency of CSMA/CD:

In the previous example we have seen that in worst case $2 T_{P}$ time require to detect a collision.

There could be many collisions may happen before a successful completion of transmission of a packet.

We are given number of collisions (contentions slots)=4.

$\text { Propagation day }=\frac{\text {distance}}{\text {speed}}$

Distance = 1km = 1000m

$\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}$

7 0

3 years ago