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3 years ago

Two positive charges of 2.8 x 10^-6 are separated by 0.50m. What force exists between the two charges?

Physics

1 answer:

Damm [24]3 years ago

6 0

F = k(Q₁Q₂)/r²
where k is Coulomb's constant 8.99e9
F = 8.99e9(2.8e-6)²/0.5² = 0.2819N

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A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s

Natasha2012 [34]

Explanation:

There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

6 0

3 years ago

What is the Milgram experiment?

lora16 [44]

Stanley Milgram, a psychologist at Yale University, conducted anexperiment<span> focusing on the conflict between obedience to authority and personal conscience. He examined justifications for acts of genocide offered by those accused at the World War II Nuremberg War Criminal trials.</span>

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3 years ago

To determine the location of the center of mass (or center of gravity) of a car, the car is drivenover a scale on a horizontal f

just olya [345]

Explanation:

It is given that,

When the front wheels are over the scale, the weight recorded by the scale is 5800 N, F₁ = 5800 N

When the rear wheels are over the scale, the scale reads 6500 N, F₂ = 6500

The distance between the front and rear wheels is measured to be 3.20 m, x₂ = 3.2 m

We need to find the location of center of mass behind the front wheels. Let the center of is located at a distance of x₁. Thus balancing the torques we get :

$5800\times x_2=6500\times (3.2-x_2)$