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sergiy2304 [10]
3 years ago
11

The truck travels in a circular path having a radius of 50 m at a speed of v = 4 m>s. For a short distance from s = 0, its sp

eed is increased by v # = (0.05s) m>s 2, where s is in meters. Determine its speed and the magnitude of its acceleration when it has moved s = 10 m.
Engineering
1 answer:
miss Akunina [59]3 years ago
6 0

Answer:

Magnitude of velocity when s is 10 m= 4.58m/s

Magnitude of acceleration when s is 10 m = 0.653 m/s²

Explanation:

Radius of circular path = 50m

Speed = 4m/s

increase in acceleration = (0.05s)m/s²

speed and acceleration when s is 10 m = ?

Magnitude of Velocity:

From third equation of motion

v_{f}^{2}-v_{i}^{2}=2aS\\\\v_{f}=\sqrt{2aS+v_{i}^{2}} \\\\v_{f}=\sqrt{2(0.05)(10)+(4)^{2}} \\\\v_{f}=4.58\,m/s

Magnitude of Acceleration:

a=\sqrt{a_{t}^{2}+a_{n}^{2}}--(1)

The tangential component of acceleration is

a_{t}=(0.05)(10)\\a_{t}=0.5\,m/s^{2}

The normal component of acceleration is

a_{n}=\frac{v^{2}}{R}\\\\a_{n}=\frac{16}{50}\\\\a_{n}=0.42\,m/s^{2}

Substituting tangential and normal component in (1)

a=\sqrt{(0.5)^{2} +(0.42)^{2} }

a=0.653 m/s²

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Answer:

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Answer:

  1. public class Main {
  2.    public static void main(String[] args) {
  3.        String testString = "abscacd";
  4.        String evenStr = "";
  5.        String oddStr = "";
  6.        for(int i=testString.length() - 1; i >= 0; i--){
  7.            if(i % 2 == 0){
  8.                evenStr += testString.charAt(i);
  9.            }
  10.            else{
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  12.            }
  13.        }
  14.        System.out.println(evenStr + oddStr);
  15.    }
  16. }

Explanation:

Firstly, let declare a variable testString to hold an input string "abscacd" (Line 1).

Next create another two String variable, evenStr and oddStr and initialize them with empty string (Line 5-6). These two variables will be used to hold the string at even index and odd index, respectively.

Next, we create a for loop that traverse the characters of the input string from the back by setting initial position index i to  testString.length() - 1  (Line 8). Within the for-loop, create if and else block to check if the current index, i is divisible by 2, (i % 2 == 0), use the current i to get the character of the testString and join it with evenStr. Otherwise, join it with oddStr (Line 10 -14).

At last, we print the concatenated evenStr and oddStr (Line 18).  

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