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Flauer [41]
3 years ago
8

In a certain velocity selector consisting of perpendicular electric and magnetic fields, the charged particles move toward the e

ast, and the magnetic field is directed to the north. What direction should the electric field point?

Physics
1 answer:
Aneli [31]3 years ago
5 0

Answer:

Down.

Explanation:

In the given question, the charged particle is moving towards the east direction. Suppose east direction towards the positive x-axis.

And the magnetic field directed towards to the north direction. Suppose north direction towards the positive y-axis.

Therefore force of magnetic field can be calculate by the formula,

F=q(vi\times Bj)\\F=qvB(k)

Therefore force of the magnetic field is in positive z-axis means in an upward direction.

Now the force on the electric field must be in the opposite direction means it is in a downward direction.

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A car moving at 50 miles per hour speeds up steadily to 70 miles per hour over a period of 30 minutes. How far did it travel dur
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Answer:

The car will travel 30 miles during the 30-minutes period of acceleration.

Explanation:

Given data :

Initial velocity = v₁ = 50 miles/hour

Final velocity = v₂ = 70 miles/hour

Time = t = 30 min = 0.5 hour

Using the definition of acceleration, we find the acceleration (a)

                   a = (v₂ - v₁) ÷ t

                   a = (70 - 50) ÷ 0.5

                   a = 20 ÷ 0.5

                   a = 40 miles/hour²

Using 3rd equation of motion, we find the distance travel (s)

                    2as = v₂² -  v₁²

                    2(40)s = 70² - 50²

                    80 × s = 4900 - 2500

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Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reac
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Answer:

Time of flight  A is greatest

Explanation:

Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.

So

H = u₁² sin²θ₁ /2g

H = u₂² sin²θ₂ /2g

H = u₃² sin²θ₃ /2g

On the basis of these equation we can write

u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃

For maximum range we can write

D = u₁² sin2θ₁ /g

1.5 D = u₂² sin2θ₂ / g

2 D =u₃² sin2θ₃ / g

1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁

1.5 = u₂ cosθ₂ /u₁ cosθ₁      ( since , u₁ sinθ₁ =u₂ sinθ₂ )

u₂ cosθ₂ >u₁ cosθ₁

u₂ sinθ₂ < u₁ sinθ₁

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Time of flight B < Time of flight  A

Similarly we can prove

Time of flight C < Time of flight B

Hence Time of flight  A is greatest .

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3 years ago
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