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Luba_88 [7]
3 years ago
15

Based on these answers how can I determine when electric field strength is affected or not

Physics
1 answer:
Stolb23 [73]3 years ago
6 0

It's not that hard. Let me explain...

There are two ways to remember this.

The first one is by using the formula for the electric strength (E)

E = k × (Q / r^2) , where:

  • k is a constant
  • Q is the "central" charge. The one producing the electric field. <u>Not</u> the one we use to test it.
  • r is the distance between the charges

As you can see, field's strength is proportional to the magnitude of Q, and inverse proportional to the square of the distance. The field does not care how much you alter the charge you positioned in ot to test it.

Although that is the only way to work out the problem, there are many ways to remember it. Given that i have difficulty in remembering formulas, i always try to derive it from something i know.

By definition: E is equal to the force acted on the test charge, divided by its value.

E = F / q

  • F --> Coulomb's force
  • q --> test charge

Since i remember that:

F = k ( Q × q) / r^2

just take out from the expression the q term, since you divided by it, and all it remains is the strength of the field.

In other words, the strength changes as long as something from the expression is altered.

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A rectangular copper strip 1.5cm wide and 0.10cn thick carries a current of 5.0A. Find the Hall voltage for a 1.2T magnetic fiel
Scrat [10]

Answer:

4.4345× 10^-7V

Explanation:

The computation of the half voltage for a 1.2T magnetic field applied is shown below

The volume of one mole of copper is

v = m ÷p

= 63.5 ÷ 8.92

= 7.12cm

Now the density of free electrons in copper is

n = Na ÷ V

= 6.02 × 10^23 ÷ 7.12

= 8.456× 10^28/m^3

Now the half voltage is

= IB ÷ nqt

= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)

= 4.4345× 10^-7V

7 0
3 years ago
Describe the energy change in the particles of a substance during melting. A) The kinetic energy of the particles remains unchan
Rina8888 [55]

Answer:

As ice melts into water, kinetic energy is being added to the particles. This causes them to be 'excited' and they break the bonds that hold them together as a solid, resulting in a change of state: solid -> liquid.

Explanation:

As we may know, the change in state of an object is due to the change in the average kinetic energy of the particles.

This average kinetic energy is proportional to the temperature of the particles.

This is because heat is a form of energy; by adding energy to ice - heat, you "excite" the water molecules, breaking the interactions in the lattice structure and forming weaker, looser hydrogen-bonding interactions.

This causes the ice to melt. This is demonstrated in the image below.

More generally, when you remove energy - the object cools down, the particles move a lot slower. So slow, that they individually attract other molecules more than before, and this results in a physical change that also changes the state.

3 0
3 years ago
Read 2 more answers
5p - 14 = 8 p + 4 find p
Digiron [165]

Answer:

Explanation:

5p - 14 = 8p + 4

5p = 8p + 18 <-- Moving constants to one side; add the same number of +14 to both sides.

-3p = 18. <-- The same thing with the variable itself.

p = -6 <-- Divide both sides by negative 3.

6 0
3 years ago
The tow spring on a car has a spring constant of 3,086 N / m and is initially stretched 18.00 cm by a 100.0 kg college student o
sdas [7]

Answer:

The velocity of the skateboard is 0.774 m/s.

Explanation:

Given that,

The spring constant of the spring, k = 3086 N/m

The spring is stretched 18 cm or 0.18 m

Mass of the student, m = 100 kg

Potential energy of the spring, P_f=20\ J

To find,

The velocity of the car.

Solution,

It is a case of conservation of energy. The total energy of the system remains conserved. So,

P_i=K_f+P_f

\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2+20

\dfrac{1}{2}\times 3086\times (0.18)^2=\dfrac{1}{2}mv^2+20

50-20=\dfrac{1}{2}mv^2

30=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{60}{100}}

v = 0.774 m/s

So, the velocity of the skateboard is 0.774 m/s.

7 0
3 years ago
find the potential energy of an aircraft weighing 10000 bs at 5000 ft true altitude and 125 kts true air speed
Cloud [144]

Answer:

U=5*10^7ft.Ib

Explanation:

From the question we are told that

Weight  W= 10000bs

Altitude H=5000ft

Speed    V=125kts\\1kts=0.514m/s\\V=125*0.514=>64.25m/s

Generally the equation for Potential energy ids mathematically given as

Potential\ Energy\ U=mgh

U=Wh

U=10000*5000

U=5*10^7ft.Ib

6 0
3 years ago
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