1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Gwar [14]
3 years ago
14

The strength of gravity on the Moon is 1.6

Computers and Technology
1 answer:
sergeinik [125]3 years ago
6 0

Answer:

Mass = 80kg

Explanation:

Given

On Earth

Mass = 80kg

On the moon

g = 1.6N/kg

Required

The astronaut's mass on the moon

The mass of an object do not change base on location

So, if the mass of the astronaut is 80kg on earth, it will be 80 kg on the moon.

Hence:

Mass = 80kg

You might be interested in
Your task is to write a C program that measures the latencies of various system calls. In particular, you want to know 1) the co
tensa zangetsu [6.8K]

Answer and Explanation:

#include <stdio.h>

#include<fcntl.h>

#include <sys/time.h>

#include<time.h>

#define MAX 1000

int main()

{

int pid;

int i,fd ;

char c[12];

FILE *fp;

struct timeval start,end;

double time1,time2,time3;

//open file for writing

fp=fopen("output.txt","w");

 

if(!fp)

{

printf("Not able to open the file output.txt\n");

return -1;

}

for(i = 0; i < MAX ; i++)

{

gettimeofday(&start,NULL);

//invoke getpid call

system(pid = getpid());

//printf("%d\n",start.tv_usec);

}

gettimeofday(&end,NULL);

//printf("%d\n",(end.tv_usec - start.tv_usec));

time1 = ((end.tv_sec - start.tv_sec) + (end.tv_usec - start.tv_usec) / 1000000.0)/MAX;

 

//wtite the time taken to execute getpid to

//to get micro second , divide multiply time by 1000000 , to get nano multiply time by 1000000000

printf("getpid(): %.10f %.10f\n",time1*1000000,time1*1000000000);

fprintf(fp,"getpid():%.10f %.10f\n",time1*1000000,time1*1000000000);

//in similar way execute other two commands ,open and read

 

for(i = 0; i < MAX ; i++)

{

gettimeofday(&start,NULL);

//invoke getpid call

system(open("/dev/null", O_RDONLY ));

//printf("%d\n",start.tv_usec);

}

gettimeofday(&end,NULL);

//printf("%d\n",(end.tv_usec - start.tv_usec));

time2 = ((end.tv_sec - start.tv_sec) + (end.tv_usec - start.tv_usec) / 1000000.0)/MAX;

 

//wtite the time taken to execute getpid to

printf("open(): %.10f %.10f\n",time2*1000000,time2*1000000000);

fprintf(fp,"open():%.10f %.10f\n",time2*1000000,time2*1000000000);

//in similar way execute other two commands ,open and read

fd = open("/dev/dev",O_RDONLY );

//printf("fd = %d\n",fd);

for(i = 0; i < MAX ; i++)

{

gettimeofday(&start,NULL);

//invoke getpid call

system( read(fd,c,10));

//printf("%d\n",start.tv_usec);

}

gettimeofday(&end,NULL);

//printf("%d\n",(end.tv_usec - start.tv_usec));

time3 = ((end.tv_sec - start.tv_sec) + (end.tv_usec - start.tv_usec) / 1000000.0)/MAX;

 

//wtite the time taken to execute getpid to

printf("read(): %.10f %.10f\n",time3*1000000,time3*1000000000);

fprintf(fp,"read(): %.10f %.10f\n",time3*1000000,time3*1000000000);

}

----------------------------------------------------------

//output

//I have written output to standard output also , you can remove that

getpid(): 0.1690000000 169.0000000000    

open(): 0.1890000000 189.0000000000    

read(): 3.1300000000 3130.0000000000

------------------------------------------------------

//Makefile content

prob2.o : prob2.c    

         gcc -c  prob2.c                                                                                                                                      

prob2 : prob2.o                                                                                                                                                

       gcc -o prob2 prob2.o                                                                                                                                    

all   :                                                                                                                                                        

       gcc -o prob2 prob2.c                                                                                                                                    

clean:                                                                                                                                                          

       rm -rf prob2.o  

---------------------------------------

use

$make all

then execute as below

$./prob2

3 0
3 years ago
Samantha was calculating a mathematical formula on an electronic spreadsheet. She used multiple values to recalculate the formul
Bezzdna [24]
<span>Random access memory. This problem requires you to know what the different types of memory are and their relative advantages and disadvantages. Let's look at them and see why 3 are wrong and one is correct. read-only memory: Otherwise known as ROM, this type of memory stores code that can't be over written. Used frequently for constant lookup values and boot code. Since it can't be written to by normal programs, it can't hold temporary values for Samantha. So this is the wrong choice. random-access memory: Otherwise known as RAM, this type of memory is used to store temporary values and program code. It is quite fast to access and most the immediately required variables and program code is stored here. It can both be written to and read from. This is the correct answer. hard disk: This is permanent long term readable and writable memory. It will retain its contents even while powered off. But accessing it is slow. Where the contents of RAM can be accessed in nanoseconds, hard disk takes milliseconds to seconds to access (millions to billions of times slower than RAM). Because it's slow, this is not the correct answer. But it's likely that Samantha will save her spreadsheet to hard disk when she's finished working with it so she can retrieve the spreadsheet later to work on again. compact disk: This is sort of the ROM equivalent to the hard disk. The data stored on a compact disk can not be over written. One way of describing the storage on a compact disk is "Write Once, Read many times". In most cases it's even slower than the hard disk. But can be useful for archiving information or making backups of the data on your computer.</span>
5 0
3 years ago
Read 2 more answers
___________ is related to mass, but also includes the gravitational pull of the Earth.
anygoal [31]
Earth's gravity comes from all its mass. All its mass makes a combined gravitational pull on all the mass in your body. That's what gives you weight. And if you were on a planet with less mass than Earth, you would weigh less than you do here. So weight is the answer
4 0
3 years ago
Given an object context for an Entity Data Model named mmaBooks, which of the following statements would you use to add a Custom
ehidna [41]

Answer:

"mmaBooks.Customers.Add(customer);" is a correct answer for the above question.

Explanation:

Missing information : The correct answer is missing in the question which is defined in the answer part.

  • If a user wants to add any objects to any collection in the C# programming, then he needs to follow the "Entity_data_model_named. collection_name. ADD(object_name)" syntax. The above question also wants this type of statement.
  • The option c states the same statements, but there is needs one statement to define the name of db or database models. But the option c does not hold the name of the database models. Hence it is not the correct answer.
  • And the other options do not follow the syntax to add, hence others is also not a valid option.
3 0
3 years ago
The code selection above is taken from the Color Sleuth activity you just completed. This selection would count as an abstractio
Vinvika [58]

Answer:

b

Explanation:

4 0
3 years ago
Other questions:
  • What is computer engineering?
    11·1 answer
  • 4. The same data source can be used multiple times in creating mail-merge documents.
    7·1 answer
  • Write a function called lucky_sevens that takes in one #parameter, a string variable named a_string. Your function #should retur
    15·1 answer
  • Students who respond promptly to e-mails are following which netiquette rule?
    13·2 answers
  • Describe the 3 different types of authentication
    6·2 answers
  • Which are two main areas of the properties inspector
    7·1 answer
  • What is the term for an understanding about the processes that underlie memory, which emerges and improves during middle childho
    7·2 answers
  • Select the correct answer.
    7·2 answers
  • Do you think people accept poor quality in information technology projects and products in exchange for faster innovation? What
    5·1 answer
  • Old systems can be useful when designing new computer software.<br> True or False
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!