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3 years ago

If a positive test charge is placed in an electric field, what is the direction of the force on the test charge?

Physics

1 answer:

svetlana [45]3 years ago

6 0

Explanation:

If a positive test charge is placed in an electric field, it will exert the force in the test charge in the direction of electric field vector. We know that the direction of electric field is given by electric field lines. The field lines for a positive charge is outwards. The electric force acting on the charge is given by :

F = q E

Hence, this is the required solution.

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A 0.299 kg mass slides on a frictionless floor with a speed of 1.44 m/s. The mass strikes and compresses a spring with a force c

VMariaS [17]

Answer:

x = 0.12 m

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

(m)×(v^2) = (k)×(x^2)

x^2 = [(m)(v^2)]/k

x = \sqrt{ [(m)(v^2)]/k}

x = \sqrt{ [(0.299)((1.44)^2)]/(44.9)}

x = 0.12 m

Therefore, the mass will travel 0.12 m and come to rest.

= 0.491 m/s

5 0

3 years ago

The repulsive force between two protons has a magnitude of 2.00 N. What is the distance between them?

Aloiza [94]

Answer:

The answer to your question is letter A. r = 1.07 x 10⁻¹⁴ m

Explanation:

Data

F = 2 N

d = ?

q = 1.6 x 10 ⁻¹⁹ C

k = 8.987 Nm²/C²

Formula

$F = K\frac{q1q2}{r^{2}}$

Solve for r

$r = \sqrt{\frac{kq1q2}{F}}$

Substitution

$r = \sqrt{\frac{8.987 x 10^{9}x1.6 x 10^{-19} x 1.6 x 10x^{-19}}{2}}$

Simplification

r = $\sqrt{\frac{2.3 x 10^{-28}}{2}}$

r = $\sqrt{1.15 x 10^{-24}}$

Result

r = 1.07 x 10⁻¹⁴ m

6 0

2 years ago

Read 2 more answers

To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you

Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

β = 10 log $\frac{I}{I_o}$

let's write this expression for our case

β₁ = 10 log \frac{I_1}{I_o}

β₂ = 10 log \frac{I_2}{I_o}

β₂ -β₁ = 10 ( $log \frac{I_2}{I_o} - log \frac{I_1}{I_o}$ )

β₂ - β₁ = 10 $log \frac{I_2}{I_1}$

log \frac{I_2}{I_1} = $\frac{60 - 20}{10}$ = 3

$\frac{I_2}{I_1}$ = 10³

I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

I = P / A

P = I A

the area is of a sphere

A = 4π r²

the power of the sound does not change, so we can write it for the two points

P = I₁ A₁ = I₂ A₂

I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

I₁ r₁² = (10³ I₁ ) r₂²

r₁² = 10³ r₂²

r₂ = r₁ / √10³

we calculate

r₂ = $\frac{10.0}{\sqrt{10^3} }$

r₂ = 0.316 m

8 0

3 years ago

If you were trying to kill a fish witha a spear, woukd you ain where you see the fish?

patriot [66]

No, you can't aim where you see the fish, because refraction will change the position of the fish and make it appear in a different position from where it actually is.

Hope this helps!

8 0

2 years ago

Read 2 more answers

Why do objects repel and attract?

Anarel [89]

Objects repel and attract because of a thing called electrostatic attraction. When objects have the same charge (positive or negative), then they will repel, and if they have opposite charges then they will attract

5 0

2 years ago