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bixtya [17]
3 years ago
4

Frequently in physics, one makes simplifying approximations. A common one in electricity is the notion of infinite charged sheet

s. This approximation is useful when a problem deals with points whose distance from a finite charged sheet is small compared to the size of the sheet. In this problem, you will look at the electric field from two finite sheets and compare it to the results for infinite sheets to get a better idea of when this approximation is valid. What is the magnitude E of the electric field at the point on thex axis with x coordinate a/2?

Physics
1 answer:
Novay_Z [31]3 years ago
7 0

Answer:

Explanation:

solution found below

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If a car is moving with a speed of 5 km/h on a highway. Find the distance travelled by the car in 2 hours?.​
grin007 [14]

Answer:

Distance = 10 kilometers.

Explanation:

Given the following data;

Speed = 5km/h

Time = 2 hours

To find the distance traveled;

Distance = speed * time

Distance = 5 * 2

Distance = 10 kilometers

Therefore, the distance travelled by the car is 10 kilometers.

5 0
3 years ago
An object–spring system undergoes simple harmonic motion. If the amplitude increases but the mass of the object is not changed,
cricket20 [7]

Answer:

The correct answer is:

doesn't change (d)

Explanation:

The total energy in a system is the sum of Kinetic and Potential energies in a system, assuming that energy is not lost to an external procedure. Now, let us define what potential and kinetic energies are:

Potential Energy: this is energy at rest or stored energy

Kinetic Energy: this is energy in motion

In a simple harmonic motion of a mass-spring system, there is no dissipative force, hence the total energy is equal to the potential and kinetic energies. The total energy is not changed rather, it varies between potential and kinetic energies depending on the point at which the mass is. The kinetic energy is greatest at the point of lowest amplitude (highest velocity) and lowest at the point of greatest amplitude (lowest velocity), while potential energy is greatest at the point of highest amplitude (lowest velocity) and lowest at the point of smallest amplitude ( highest velocity). However, at every point, the sum of kinetic and potential energies equals total energy.

6 0
3 years ago
When was the first airplane made?
expeople1 [14]
The first ariplanr was made December 17, 1903
3 0
4 years ago
Read 2 more answers
The current theory of the structure of the
IRISSAK [1]

1) The mass of the continent is 3.3\cdot 10^{21} kg

2) The kinetic energy of the continent is 624 J

3) The speed of the jogger must be 4 m/s

Explanation:

1)

We start by finding the volume of the continent. We have:

L = 5850 km = 5.85\cdot 10^6 m is the side

t = 35 km = 3.5\cdot 10^4 m is the depth

So the volume is

V=L^2 t = (5.85\cdot 10^6)^2 (3.5\cdot 10^4)=1.20\cdot 10^{18} m^3

We also know that its density is

d=2750 kg/m^3

Therefore, we can find the mass by multiplying volume by density:

m=dV=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg

2)

The kinetic energy of the continent is given by:

K=\frac{1}{2}mv^2

where

m=3.3\cdot 10^{21} kg is its mass

v = 3.2 cm/year is the speed

We have to convert the speed into m/s. We have:

3.2 cm = 0.032 m

1 year = 1(365)(24)(60)(60)=3.15\cdot 10^7 s

So, the speed is:

v=\frac{0.032 m}{3.15 \cdot 10^7 s}=1.02\cdot 10^{-9} m/s

So, we can now find the kinetic energy:

K=\frac{1}{2}(1.20\cdot 10^{21})(1.02\cdot 10^{-9})^2=624 J

3)

Here we have a jogger of mass

m = 78 kg

And the jogger has the same kinetic energy of the continent, so

K = 624 J

The kinetic energy of the jogger is given by

K=\frac{1}{2}mv^2

where v is the speed of the jogger.

Solving for v, we find the speed that the jogger must have:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(624)}{78}}=4 m/s

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0
3 years ago
The storage coefficient of a confined aquifer is 6.8x10-4 determined by a pumping test. The thickness of the aquifer is 50 m and
Ivanshal [37]

Answer

given,

storage coefficient, S = 6.8 x 10⁻⁴

thickness of aquifer, t = 50 m

porosity of the aquifer, n = 25 % = 0.25

Density of the water, γ = 9810 N/m³

Compressibilty  of water,β = 4.673 x 10⁻¹⁰ m²/N

We know,

   S = γ t(nβ + α)

where, α is the compressibility of the aquifer

   6.8 x 10⁻⁴  =9810 x 50 x (0.25 x 4.673 x 10⁻¹⁰+ α)

     α = 1.269 x 10⁻⁹ m²/N

Expansability of water

            = n t β γ

            = 0.25 x 50 x 4.673 x 10⁻¹⁰ x 9810

            = 5.73 x 10⁻⁵

7 0
3 years ago
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