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RSB [31]
3 years ago
11

When implementing a patch, it is recommended that there be an back-out strategy in place; this is necessary because it is possib

le the patch might create complications.
Engineering
1 answer:
melomori [17]3 years ago
4 0

Answer:

i dont know whats the context please

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Is normally a large red cable connected to the battery
Igoryamba

Answer:

yes

Explanation:

its the ground cable

4 0
3 years ago
Read 2 more answers
Suppose the loop is moving toward the solenoid (to the right). Will current flow through the loop down the front, up the front,
Tems11 [23]

Answer:

See explanation

Explanation:

The magnetic force is

F = qvB sin θ

We see that sin θ = 1, since the angle between the velocity and the direction of the field is 90º. Entering the other given quantities yields

F

=

(

20

×

10

−

9

C

)

(

10

m/s

)

(

5

×

10

−

5

T

)

=

1

×

10

−

11

(

C

⋅

m/s

)

(

N

C

⋅

m/s

)

=

1

×

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11

N

6 0
3 years ago
Read 2 more answers
Two particles have a mass of 7.8 kg and 11.4 kg , respectively. A. If they are 800 mm apart, determine the force of gravity acti
aleksley [76]

Answer:

A) About 9.273 \times 10^{-9} newtons

B) 76.518 newtons

C) 111.834 newtons

Explanation:

A) F_g=\dfrac{GM_1M_2}{r^2} , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:

F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}

B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.

C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.

Hope this helps!

3 0
3 years ago
A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco
allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

d_1=50 mm,f_1=0.0048

For pipe 2

d_2=75 mm,f_2=0.0058

Q=2.8 l/s

Q=2.8\times 10^{-3]

We know that Q=AV

Q=A_1V_1=A_2V_2

A_1=1.95\times 10^{-3}m^2

A_2=4.38\times 10^{-3} m^2

So V_2=0.63 m/s,V_1=1.43 m/s

head loss (h)

h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}

Now putting the all values

h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0
3 years ago
Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user
Thepotemich [5.8K]

Answer:

import java.util.Scanner;

public class FindMatchValue {

  public static void main (String [] args) {

     Scanner scnr = new Scanner(System.in);

     final int NUM_VALS = 4;

     int[] userValues = new int[NUM_VALS];

     int i;

     int matchValue;

     int numMatches = -99; // Assign numMatches with 0 before your for loop

     matchValue = scnr.nextInt();

     for (i = 0; i < userValues.length; ++i) {

        userValues[i] = scnr.nextInt();

     }

     /* Your solution goes here */

         numMatches = 0;

     for (i = 0; i < userValues.length; ++i) {

        if(userValues[i] == matchValue) {

                       numMatches++;

                }

     }

     System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

  }

}

8 0
3 years ago
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