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k0ka [10]
2 years ago
8

A uniform electric field is pointing in x direction. The magnitude of the electric field is 10 N/C. The filed makes an angle of

30 deg with a rectangular surface of area 2 x10 m Calculate the electric flux crossing the surface.
Physics
1 answer:
Ahat [919]2 years ago
8 0

Answer:

- 100 Nm²/C

Explanation:

E = magnitude of electric field along x-direction = 10 N/C

θ = angle made by the direction of electric field with rectangular surface = 30

φ = angle made by the direction of electric field with normal to the rectangular surface = 90 + 30 = 120

A = area of the rectangular surface = 2 x 10 = 20 m²

Φ = Electric flux crossing the surface

Electric flux is given as

Φ = E A Cosφ

Φ = (10) (20) Cos120

Φ = - 100 Nm²/C

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calculating displacement.

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It's not true that displacement and distance would be the same always. Displacement is always smaller than or equal to distance as it is the smallest path between the initial and final point whereas distance is the measure of the total path covered.

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A camcorder has a power rating of 15 watts. If the output voltage from its battery is 5 volts, what current does it use?
MrMuchimi

Formula

W = E * I

Givens

E = 5 volts

W = 15 watts

I = ?

Solution

W = E * I

15 = 5 * I

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3 years ago
A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin
Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0
3 years ago
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