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3 years ago

A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?

Physics

1 answer:

Andru [333]3 years ago

3 0

Answer:

F = 37.8 × 10^(6) N

Explanation:

The charges are 0.06 C and 0.07 C.

Thus;

Charge 1; q1 = 0.06 C

Charge 2; q2 = 0.07 C

Distance between them; r = 3 m

Formula for the force in between them is;

F = kq1•q2/r²

Where k is a constant = 9 × 10^(9) N.m²/C²

Thus;

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

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3 years ago

What can one say for sure about element #99, Einsteinium (Es))? a) There's no such element! b) It is an alkali metal. c) It is r

Dmitrij [34]

Answer:

Option (C)

Explanation:

Einsteinium is an element of the periodic table grouped in the Actinide series, with atomic number 99. They are dense element and highly electro-positive. <u>They are highly radioactive</u>, i.e the atoms within the element are unstable and constantly decay until they reach a stable environment. It has 99 number of electrons and protons, 153 number of neutrons.

Due to its high radioactivity, they are health hazardous and can used in making nuclear weapons but their uses are very limited and unknown.

Thus, the correct answer is option (C).

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3 years ago

Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5

schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

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3 years ago

Explain what happens to the energy in a group in a system if one object loses energy according to the Law of Conservation of Ene

konstantin123 [22]

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3 years ago

A 782.10 kg car is brought from 7.60 m/s to 3.61 m/s over a time period of 4.23 seconds. What is the average force the car exper

alexandr402 [8]

Answer:

–735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 782.10 kg

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Force (F) =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Acceleration (a) =?

a = (v – u) / t

a = (3.61 – 7.60) / 4.23

a = –3.99 / 4.23

a = –0.94 m/s²

Finally, we shall determine the force experienced by the car as shown below:

Mass (m) of car = 782.10 kg

Acceleration (a) = –0.94 m/s²

Force (F) =?

F = ma

F = 782.10 × –0.94

F = –735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

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3 years ago