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3 years ago

the electric current in a wire is 1.5 A. How many electrons flow past a given point in a time of 2 s?

Physics

1 answer:

34kurt3 years ago

6 0

Answer:

The quantity of electrons that flows past a given point is 3.0 C.

Explanation:

An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).

i.e I = $\frac{Q}{t}$

It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.

Given that; I = 1.5A and t = 2s, find Q.

Q = It

= 1.5 × 2

= 3.0 C

The quantity of electrons that flows past a given point is 3.0 C.

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Points A, B, and C lie along a line from left to right, respectively. Point B is at a lower electric potential than point A. Poi

djyliett [7]

Answer:

A) The particle will accelerate in the direction of point C.

Explanation:

As we know that

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V_A>V_B>V_C

And we know that flow is always from higher to lower potential (for positive charge due to positive potential energy).

So the charge will accelerate from B toward C.

Hence, the correct option is A.

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3 years ago

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3 years ago

A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w

professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

$\theta=70\°$ The angle at which the ball leaves the bat

$V_{o}=55 m/s$ The initial velocity of the ball

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: $x$

Firstly we need to know this velocity has two components:

Horizontally:

$V_{ox}=V_{o}cos \theta$ (1)

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

Vertically:

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (5)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :