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LenKa [72]
2 years ago
9

How is Electric current able to flow?

Physics
2 answers:
brilliants [131]2 years ago
3 0
Electric potential on one side of the closed circuit is applied and this causes the electrons to move to a position of lower potential. Imagine this as when you raise an object above the ground and release it. It moves from its higher potential energy position above the ground to its lower potential energy position.
AlekseyPX2 years ago
3 0

Explanation:

When the potential difference is provided to the conductor then the electrons start moving in the conductor. The electrons move from lower potential to the higher potential.

The rate of flow electrons constitutes the electric current. The direction of the electric current is opposite to the direction of the electrons.

For example, suppose the battery is connected to the circuit. It forces the electrons to flow in the conductor. The battery provides potential to the electrons so that the electrons can flow in the circuit. There are two terminals in the battery: Positive terminal and Negative terminal. Positive terminal is at higher potential. Negative terminal is at lower potential.

In this way, electrons flow in the circuit and constitute the electric current flow in the circuit.

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1- A 30 gram bullet travels at 300 m/s. How much kinetic energy does it have?
labwork [276]

Answer:

1.35 kJ  

Explanation:

KE = ½mv² = ½ × 0.030  kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ

7 0
3 years ago
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Clarissa thought that learning to play the banjo was a waste of time. She didn't bother to practice after her lessons
emmasim [6.3K]

Answer:

The answer ro this question is fear of failure

7 0
3 years ago
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DETERMINE THE BCD EQUIVALENT OF 1100111110101001
Solnce55 [7]

Answer: It states that the BCD equivalent would be 0001000100000000000100010001000100010000000100000001000000000001.

6 0
2 years ago
"A short-wave radio antenna is supported by two guy wires, 150 ft and 170 ft long. Each wire is attached to the top of the anten
Nonamiya [84]

Answer:

The anchor points are 78.37 ft and 111.99 ft

Explanation:

If you look at the attached (Fig 1) you will find that the union of antenna and its guy wires forms two right triangles. To solve problems that involve this kind of triangles, you can apply trigonometric functions (sine, cosine, etc) and Pythagoras Theorem. Trigonometric functions states the relation between angles, sides and hypotenuse of a right triangle. If you look Fig2, considering α angle, "b" is the opposite side, "a" the adjacent side and "c" the hypotenuse. Then

a) Sine (α) = b/c it means opposite side/hypotenuse

b) Cosine (α)= a/c, it means adjacent side/hypotenuse

c) Tangent (α) = b/a opposite side/adjacent side.

Pythagoras theorem states that if you called "a" and "b", the sides of the right triangle, and "c" the hypotenuse, then:

                                      a² + b² = c²    

As the problem states the lengths 150 ft and 170 ft represents the value of the hypotenuse of each triangle and 65° is one of the angles of the triangle with 150 ft hypotenuse. So you can solve this using sin (65°) to find the height of the antenna (h) and then the two distances (x and y,).

Sine (65°) = h/ 150 ft ⇒ Sine (65°) x 150ft = h ⇒ h = 127.9 ft.

To find x : Cosine (65°) = x/ 150 ft ⇒ Cosine (65°) x 150 ft = x

⇒ x = 78.37 ft.

And finally, to find y we can apply Pythagoras theorem

(170 ft)² = (127.9 ft)² + y² ⇒ y² = (170 ft)² - (127.9 ft)² ⇒ y = 111.99 ft

Summarizing, the anchor points are 78.37 ft and 111.99 ft

4 0
3 years ago
A first order reaction, A -> products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,
tamaranim1 [39]

Answer: a)  The rate constant, k, for this reaction is 0.00516s^{-1}

b) No t_{\frac{1}{2}} does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

A\rightarrow products

Given: Order with respect to A = 1

Thus rate law is:

a) Rate=k[A]^1

k= rate constant

0.00250=k[0.484]^1

k=0.00516s^{-1}

The rate constant, k, for this reaction is 0.00516s^{-1}

b) Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}

t_{\frac{1}{2}}=\frac{0.69}{k}

Thus t_{\frac{1}{2}} does not depend on concentration.

8 0
3 years ago
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