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Anika [276]
2 years ago
13

) The square plates of a 5000-pF parallel-plate capacitor measure 50 mm by 50 mm and are separated by a dielectric that is 0.23

mm thickand totally fills the region between the plates. The voltage rating (the maximum safe voltage) of the capacitor is 400 V. What is the maximum energy that can be stored in this capacitor without damaging it
Physics
1 answer:
notsponge [240]2 years ago
6 0

Answer:

4 x 10⁻⁴ J

Explanation:

C = 5000 pF, V = 400 V

Energy = CV²/2 = 5000 x 10⁻¹² x 400²/2 = 4 x 10⁻⁴ J

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a bicyclist travels with an averahe velocity of 15km/h north, for 20 minutes .what is his displacement ?​
Lunna [17]

Explanation:

velocity = 15 km/hr = 15×1000m / 60×60s

= 4.17 m/s

time= 20 min= 20 × 60s

= 1200s

displacement =?

we know that,

V = D / t

or, 4.17 = D / 1200

or, D = 4.17 × 1200

or,D = 5004 m

7 0
3 years ago
It is the highest velocity attainable by an object as it falls through the air.
Montano1993 [528]
<span>The highest velocity attainable by an object as it falls through the air is:

The terminal velocity</span>
8 0
3 years ago
4.A 200 kg cannon fires a 5 kg chicken into the air at a velocity of 100
exis [7]

Answer:

-2.5

Explanation:

5kg in grams = 5000

200kg in grams = 200,000

V2 = - (M1/M2) V1

Vc = - (Mh/Mc) Vh

Vc = - (5000/200000) 100

Vc = -2.5 units/second

4 0
2 years ago
An Apple with the mass of 200g falls from the tree.what is the acceleration of the apple towards the earth.what is the accelerat
notsponge [240]

Answer:

Assume that the earth is a sphere (with radius equal to its actual radius at the equator) of uniform density.

  • Acceleration of the apple towards the earth: approximately 9.79\; \rm m \cdot s^{-2}.
  • Acceleration of the earth towards the apple: approximately 3.28 \times 10^{-25}\; \rm m \cdot s^{-2}.

Both values were rounded to three significant figures. Air resistance is assumed to be negligible.  

Explanation:

Convert the mass of the apple to standard units:

m(\text{apple}) = 200\; \rm g = 0.200\; \rm kg.

Look up the mass and radius of the earth:

  • Mass of the earth: m(\text{earth})\approx 5.97\times 10^{24}\; \rm kg.
  • Radius of the earth: r \approx 6.3781\times 10^{6}\; \rm m (at the equator.)

Assume that the earth is a sphere of uniform density. The acceleration of an object moving towards the earth under free fall would be approximately equal to the gravitational field strength of the earth at that position.

For a sphere of mass m (assuming uniform density,) the gravitational field strength g at a distance of r away from the center of the sphere would be:

\displaystyle g = \frac{G \cdot m}{r^2}.

The height of the tree should be much smaller than the radius of the earth. Therefore, the distance between the apple and the center of the earth would be approximately equal to the radius of the earth.

The strength of the gravitational field of the earth at the position of the apple would be approximately:

\begin{aligned} & \frac{G \cdot m(\text{earth})}{r^2} \\ &\approx \frac{6.67\times 10^{-11}\; \rm m^3\cdot kg^{-1}\cdot s^{-2} \times 5.97 \times 10^{24}\; \rm kg}{\left(6.3781\times 10^{6}\; \rm m\right)^2} \\ &\approx 9.79\; \rm m\cdot s^{-2} \end{aligned}.

This value should be approximately equal to the acceleration of the apple towards the earth.

On the other hand, assume that the apple acts like a point mass when compared to the earth. In other words, assume that the radius of the apple is much smaller than the distance between the apple and the earth.

It can be shown (using calculus) that if the earth is a sphere with uniform density, the earth will behave just like a point mass at the center of the earth when studying interactions between the earth and objects outside of it.

At the center of the earth, the strength of the gravitational field due to the apple would be:

\begin{aligned} & \frac{G \cdot m(\text{apple})}{r^2} \\ &\approx \frac{6.67\times 10^{-11}\; \rm m^3\cdot kg^{-1}\cdot s^{-2} \times 0.200\; \rm kg}{\left(6.3781\times 10^{6}\; \rm m\right)^2} \\ &\approx 3.28 \times 10^{-25}\; \rm m\cdot s^{-2} \end{aligned}.

Under the assumption that the earth is a sphere of uniform density, this value should be equal to the acceleration of the earth towards the apple due to the gravitational attraction of the apple.

6 0
3 years ago
A 0.500 m length of wire with a cross-sectional area of 3.14 * 10 ^ - 6 meters squared is found to have a resistance of 2.53 * 1
mr_godi [17]

Answer:

The wire is made of silver (ρ =  1.59×10⁻⁸ ohms/m)

Explanation:

Applying,

R = ρL/A................. Equation 1

Where R = Resistance length of the wire, ρ = Resistivity of the wire, L = Length of the wire, A = crosssectional area of the wire

make ρ the subject of the equation

ρ = RA/L............. Equation 2

From the question,

Given: R = 2.53×10⁻³ ohms, A = 3.14×10⁻⁶ m², L = 0.5 m

Substitute these values into equation 2

ρ = (2.53×10⁻³)(3.14×10⁻⁶)/0.5

ρ = 1.59×10⁻⁸ ohms/m

Hence from the resistivity chart, the wire is made of silver

7 0
3 years ago
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