What is the value of the magicPowers variable after executing the following code snippet? String magicPowers = ""; int experienc
eLevel = 9; if (experienceLevel > 10) { magicPowers = magicPowers + "Golden sword "; } if (experienceLevel > 8) { magicPowers = magicPowers + "Shining lantern "; } if (experienceLevel > 2) { magicPowers = magicPowers + "Magic beans "; }
1 answer:
Answer:
Shining lantern Magic beans
Explanation:
I will explain the code line by line.
The first statement contains a string type variable magicPowers and second statement has an int type variable experienceLevel which is initialized by value 9.
Nested if statement are used and lets see which of those statements execute.
The first if statement checks if the value of experienceLevel is greater than 10. This statement evaluates to false because the value of experienceLevel is set to 9. Then the program control moves to the next if statement.
The second if statement checks if the value of experienceLevel is greater than 8. This if condition evaluates to true because the value of experienceLevel is 9. So this if statement is executed which contains the statement magicPowers = magicPowers + "Shining lantern "; . This means magicPowers stores the string Shining lantern.
The third if statement checks if the value of experienceLevel is greater than 2. As the value of experienceLevel is 9 so this condition also evaluates to true which means the statement magicPowers = magicPowers + "Magic beans "; is executed. This statement means magicPowers stores the string Magic Beans
You can use a print statement to display the output on the screen as:
System.out.print(magicPowers);
Output:
Shining lantern Magic beans
The program along with the output is attached as a screenshot.
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Answer:
V_t=6 = 32 m/s
Explanation:
The origin is at point 0 with the positive motion to the right
The instantaneous acceleration is change of velocity measured at infinitesimal interval of time, so the expression for instantaneous acceleration is:
a=dv/dt
From here we can express dv as:
dv = a dt
Replace a by 2t — 1
dv = (2t — 1) dt
Integrate both sides of equation
v=t
a=t_0
putting these value in integral
<em>v-v_0=(t^2-t)-(t_0^2-t_0)</em>
We know that v_0 = 2 at t_0 = 0, so we'll replace t_0 and v_0 by their values
v — 2 = (t^2 — t) — (0^2 — 0)
From here we can write the expression for v as:
v_t=6=6^2-6+2 (1)
So the velocity at t = 6 s is:
v_t=6 = 32 m/s
V_t=6 = 32 m/s
In order to determine the total distance travelled, we must check how maw times the particle has changed its direction, i.e. how many times its speed was equal to zero
To do that, we'll just replace v by 0 in expression (1)
0 = t^2 — t + 2
The roots of the quadratic equation are:
t_1/2=1± √(1^2-4*2*1)/2
Since 1^2-4*2*1 < 0, the quadratic equation have no real roots, so we can say that the velocity is always positive, i.e. to the right
Now that we have all the details, we can correctly draw the path of the particle
We can see from the sketch that the total distance traveled is:
s^T=Δs_0-1
s^T=| s_1 - s_0 |
Replace s_0 by its value
s^T=| s_1 - 1 | (2)
In order to determine the position of particle at t = 6 s, we'll need to determine the expression for s as function of time
Since we have already wrote expression for v as function of time (step 2), we'll use expression to get the expression for s
v= ds/dt
Multiply both sides of equation by dt
v dt = ds
Replace v by expression (1)
(t^2 — t + 2) dt = ds
Integrate both sides of equation
t=s
b=(s=0)
x=(t^2 — t + 2)
dx=ds
putting these value in integral
(t^3/3-t^2/2+t)-(t_0^3/3-t_0^2/2+t_0)= s-s_0
Since s = 1 m at t = 0, and we want to determine the position s at t = 6, we'll replace so by 1, t_0 by 0 and t by 6
(6^3/3-6^2/2+6)-(0^3/3-0^2/2+0)=s_t=6-1
