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prisoha [69]
3 years ago
13

..................................................

Engineering
2 answers:
34kurt3 years ago
6 0

Answer: yayyyy free sh.it

Explanation:

Scorpion4ik [409]3 years ago
3 0
..................................................
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A 14-lb crate is pulled up a frictionless 40° ramp with an initial velocity of v1=0.4 ft/s. It is pulled 0.3 ft from location #1
Morgarella [4.7K]

Answer:

3.25 ft/s

Explanation:

The crate is of =14-lb=m₁

The angle of inclination is = 40°=Ф

The initial velocity = 0.4 ft/s= v₁

Distance the crate will move is= 0.3 ft =d

The load pulling downwards is = 36 lb= m₂

Acceleration of the pulley, a= m₂g - m₁gsinФ / m₁+m₂ where g= 32.17 ft/s^2

a= 36*32.17 - 14*32.17*sin 40° / 14+36

a=17.37 ft/s^2

Apply the formula for final velocity

V₂²=V₁²+2ad

V₂²=0.4²+ 2*17.37*0.3

V₂²=10.582

V₂ =√10.582 = 3.25 ft/s

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3 years ago
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3 years ago
I don’t understand this
blondinia [14]

Answer:

Sorry for the delayed response- Right now I don't have time to give you the answer, but I really want to help so I'll try to phrase it in a easier way to understand things: Basically what you need to do for this problem is find the area of the base of the figure (which means length x width) and then you would simply find the volume of 160cm^{2} by finding the length of each side of the figure, find the length of the figure, find the height of the figure and then find the radius.

Have an amazing day and I hope this can somewhat help :)

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The answer is here......

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import java.util.*;

public class BarChart

{

public static void main(String args[])

{

int arr[]=new int[5];

Scanner sc=new Scanner(System.in);

for(int i=0;i<5;i++)

{

while(true){

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arr[i]=sc.nextInt();  

if(arr[i]>0)

break;

}

}

System.out.println("SALES BAR CHART");

for(int i=0;i<5;i++)

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System.out.println("Store "+(i+1)+": ");

for(int j=0;j<arr[i];j=j+100)

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System.out.print("*");

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System.out.println("");

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