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3 years ago

You drop a rock down a well that is 11.5 m deep.How long does it take the rock to hit the bottom of the well ?

1 answer:

4 0

The acceleration of gravity depends on the distance from
the center of the Earth. We use 9.8 m/s² for any place that's
on the Earth's surface, or reasonably close. The 11.5 meters
down into the well certainly qualifies.

The formula we want is:

Distance of fall = (1/2) · (acceleration) · (time)²

      11.5 m = (1/2) (9.8 m/s²) (time)²

Divide each side
by 4.9 m/s² : (11.5m) / (4.9 m/s²) = time²

   2.347 sec² = time²

   time = √(2.347 sec²) =   1.53 seconds .

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3 years ago

How much heat will be needed to warm 187 grams of water from 10 0C to 90 0C?

kvv77 [185]

<h3>Hello there!</h3>

Here, you are looking for the amount of heat put in for water, at a mass of 187 grams, to change by 80 degrees.

The equation commonly accepted to find the answer to questions like these is the specific heat formula.

The equation is Q = mc∆T, where Q is the amount of energy put in to raise the temperature by a certain amount, m is the mass, c is the specific heat capacity, and ΔT is the amount of temperature change.

The information given:

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Now just plug everything in to solve.

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3 years ago

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larisa86 [58]

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3 years ago

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What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

konstantin123 [22]

Answer:

$I=2.71\times 10^{-5}\ A$

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

$C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}$ , r is radius

Let I is the displacement current. It is given by :

$I=C\dfrac{dV}{dt}$

Here, $\dfrac{dV}{dt}$ is rate of increasing potential difference

$I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A$

So, the value of displacement current is $2.71\times 10^{-5}\ A$ .

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