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3 years ago

What is the opposite of 4/4

Physics

1 answer:

jenyasd209 [6]3 years ago

6 0

Answer: -1 or -4/4

Explanation:

4/4 simplifies to 1 and the opposite of 1 is -1.

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1. Driving at slower speeds than traffic flow _____________ .

r-ruslan [8.4K]

The correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge.

Further Explanation:

The electric field intensity at a point is the measure of the force exerted by a charge particle on another charge particle in the particular area of its strength.

The electric field intensity at a distance $d$ due to a static charge having charge $q$ is directly proportional to the amount of charge and inversely proportional to the square of the distance between them.

The Electric field intensity due to a charge is given as:

$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}$

Here, $E$ is the electric field intensity, $q$ is the amount of charge and $r$ is the distance of the charge from the point.

The above expression of electric field shows that the electric field intensity at a point depends on the amount of charge as well as the distance of the point from the charge.

Thus, the correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge.

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Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Electrostatics

Keywords: Strength, electric field, charge, distance, electric field intensity, magnitude of charge, electrostatic, test charge, kq/r^2.

7 0

4 years ago

Read 2 more answers

Particles q1 = -53.0 uc, q2 = +105 uc, and

Nimfa-mama [501]

Answer:

$-180.38\ \text{N}$

Explanation:

$q_1=-53\ \mu\text{C}$

$q_2=105\ \mu\text{C}$

$q_3=-88\ \mu\text{C}$

r = Distance between the charges

$r_{12}=0.5\ \text{m}$

$r_{23}=0.95\ \text{m}$

$r_{13}=1.45\ \text{m}$

k = Coulomb constant = $9\times 10^9\ \text{Nm}^2/\text{C}^2$

Net force is given by

$F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}$