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Gennadij [26K]
2 years ago
5

A labor push a crate through 12m and 720 J of work .The magnitude of force was

Physics
1 answer:
Harrizon [31]2 years ago
6 0

Answer:

<h2>60 N</h2>

Explanation:

The magnitude of the force can be found by using the formula

f =  \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{720}{12}  = 60 \\

We have the final answer as

<h3>60 N</h3>

Hope this helps you

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A mixture that forms when one substance dissolves another is called
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Answer:

a solution

Explanation: A solution is a homogeneous mixture consisting of a solute dissolved into a solvent .

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Please help me,its urgent!!
GREYUIT [131]

Test:

Performing a Litmus Test

Result:

Litmus paper gives the user a general indication of acidity or alkalinity as it correlates to the shade of red or blue that the paper turns.

  • To test the pH of a substance, dip a strip of litmus paper into the solution or use a dropper or pipette to drip a small amount of solution onto the litmus paper.
  • Blue litmus paper can indicate an acid with a pH between 4 and 5 or lower.
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3 years ago
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

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If they become closer, it is increased, and if the objects become farther away is decreased.
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