Write an interactive program to calculate the volume and surface area of a three-dimensional object.
Answers
Answer:
I am writing a Python program:
pi=3.14 #the value of pi is set to 3.14
radius = float(input("Enter the radius of sphere: ")) #prompts user to enter the radius of sphere
surface_area = 4 * pi * radius **2 #computes surface area of sphere
volume = (4.0/3.0) * (pi * radius ** 3) #computes volume of sphere
print("Surface Area of sphere is: ", surface_area) #prints the surface area
print("Volume of sphere is: {:.2f}".format(volume)) #prints the volume
Explanation:
Please see the attachment for the complete interactive code.
The first three lines of print statement in the attached image give a brief description of the program.
Next the program assigns 3.14 as the value of π (pi). Then the program prompts the user to enter the radius. Then it computes the surface area and volume of sphere using the formulas. Then it prints the resultant values of surface area and volume on output screen.
The psuedocode for this program is given below:
PROGRAM SphereVolume
pi=3.14
NUMBER radius, surface_area, volume
INPUT radius
COMPUTE volume = (4/3) * pi* radius ^3
COMPUTE surface_area = 4 * pi * radius ^ 2
OUTPUT volume , surface_area
END
Related Questions
A machine on a 10 Mbps network is regulated by a token bucket algorithm with a fill rate of 3 Mbps. The bucket is initially filled to capacity at 3MB. How long can the machine transmit at the full 10 Mbps capacity
Answers
Consider the following calling sequences and assuming that dynamic scoping is used, what variables are visible during execution of the last function called? Include with each visible variable the name of the function in which it was defined.a. Main calls fun1; fun1 calls fun2; fun2 calls fun3b. Main calls fun1; fun1 calls fun3c. Main calls fun2; fun2 calls fun3; fun3 calls fun1d. Main calls fun3; fun3 calls fun1e. Main calls fun1; fun1 calls fun3; fun3 calls fun2f. Main calls fun3; fun3 calls fun2; fun2 calls fun1void fun1(void);void fun2(void);void fun3(void);void main() {Int a,b,c;…}void fun1(void){Int b,c,d;…}void fun2(void){Int c,d,e;…}void fun3(void){Int d,e,f;…}
Answers
Answer:
In dynamic scoping the current block is searched by the compiler and then all calling functions consecutively e.g. if a function a() calls a separately defined function b() then b() does have access to the local variables of a(). The visible variables with the name of the function in which it was defined are given below.
Explanation:
In main() function three integer type variables are declared: a,b,c
In fun1() three int type variables are declared/defined: b,c,d
In fun2() three int type variables are declared/defined: c,d,e
In fun3() three int type variables are declared/defined: d,e,f
a. Main calls fun1; fun1 calls fun2; fun2 calls fun3
Here the main() calls fun1() which calls fun2() and fun2() calls func3() . This means first the func3() executes, then fun2(), then fun1() and last main()
Visible Variable: d, e, f Defined in: fun3
Visible Variable: c Defined in: fun2 (the variables d and e of fun2
are not visible)
Visible Variable: b Defined in: fun1 ( c and d of func1 are hidden)
Visible Variable: a Defined in: main (b,c are hidden)
b. Main calls fun1; fun1 calls fun3
Here the main() calls fun1, fun1 calls fun3. This means the body of fun3 executes first, then of fun1 and then in last, of main()
Visible Variable: d, e, f Defined in: fun3
Visible Variable: b, c Defined in: fun1 (d not visible)
Visible Variable: a Defined in: main ( b and c not visible)
c. Main calls fun2; fun2 calls fun3; fun3 calls fun1
Here the main() calls fun2, fun2 calls fun3 and fun3 calls fun1. This means the body of fun1 executes first, then of fun3, then fun2 and in last, of main()
Visible Variable: b, c, d Defined in: fun1
Visible Variable: e, f Defined in: fun3 ( d not visible)
Visible Variable: a Defined in: main ( b and c not visible)
Here variables c, d and e of fun2 are not visible
d. Main calls fun3; fun3 calls fun1
Here the main() calls fun3, fun3 calls fun1. This means the body of fun1 executes first, then of fun3 and then in last, of main()
Visible Variable: b, c, d Defined in: fun1
Visible Variable: e, f Defined in: fun3 ( d not visible )
Visible Variable: a Defined in: main (b and c not visible)
e. Main calls fun1; fun1 calls fun3; fun3 calls fun2
Here the main() calls fun1, fun1 calls fun3 and fun3 calls fun2. This means the body of fun2 executes first, then of fun3, then of fun1 and then in last, of main()
Visible Variable: c, d, e Defined in: fun2
Visible Variable: f Defined in: fun3 ( d and e not visible)
Visible Variable: b Defined in: fun1 ( c and d not visible)
Visible Variable: a Defined in: main ( b and c not visible)
f. Main calls fun3; fun3 calls fun2; fun2 calls fun1
Here the main() calls fun3, fun3 calls fun2 and fun2 calls fun1. This means the body of fun1 executes first, then of fun2, then of fun3 and then in last, of main()
Visible Variable: b, c, d Defined in: fun1
Visible Variable: e Defined in: fun2
Visible Variable: f Defined in: fun3
Visible Variable: a Defined in: main
Cloud computing gives you the ability to expand and reduce resources according to your specific service requirement.
a. True
b. False
Answers
Answer:
a. True
Explanation:
Cloud computing can be defined as a type of computing that requires shared computing resources such as cloud storage (data storage), servers, computer power, and software over the internet rather than local servers and hard drives.
Generally, cloud computing offers individuals and businesses a fast, effective and efficient way of providing services.
In Computer science, one of the most essential characteristics or advantages of cloud computing is rapid elasticity.
By rapid elasticity, it simply means that cloud computing gives you the ability to expand and reduce resources according to your specific service requirement because resources such as servers can be used to execute a particular task and after completion, these resources can then be released or reduced.
Some of the examples of cloud computing are Google Slides, Google Drive, Dropbox, OneDrive etc.
