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Lyrx [107]
3 years ago
6

What is approximate resistance between P and Q? a) 0.5 b)0.8 c)2.0 d)2.2 e)3.6

Physics
1 answer:
Rufina [12.5K]3 years ago
8 0

Answer:

0.5 Ohms

Explanation:

We note that the node Q is also between the resistors of 1ohm and 2ohms.

We note that the node P is also between the resistors of 2ohms and 3ohms.

Thus, all these resistors are in parallel, beween nodes P and Q

1/Re=1/R1+1/R2+1/R3

1/Re=1/1+1/2+1/3=(6+3+2)/6=11/6 [ohm^(-1)]

Re=6/11=0.54ohms

Rounding to the tenth: Re=0.5 ohms

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Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magni
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Answer:

a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate,  c)  x = 6.19 10⁹ m

Explanation:

This exercise looks at the motion of a positively charged particle in an electric field.

a) Since the field is vertical the acceleration in this direction is

            F = m a

the electric force is

           F = q E

we substitute

          q E = m a

           a = qE / m

the mass of the particle is m = 2.00 10-16 kg

           a = 1.6 10⁻¹⁹ 2.02 10³ / 2.00 10⁻¹⁶ kg

           a = 1,616 m / s²

           

on the x-axis there are no relationships because there are no forces.

Since the particle has velocities in both axes, its motion is PARABOLIC,

b) the positive particle is accelerated towards the negative plate,

The field is descending, for which the event is down

c) where  hit the particle on the x-axis

they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.

Let's find the components of the initial velocity.

           sin θ = v_{oy} / v

           cos θ = v₀ₓ / v

           v_{oy} = v₀ sin θ

           v₀ₓ = v₀ cos θ

           v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s

           v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s

Let's find the time it takes to hit the negative plate

            y = y₀I + v_{oy} t + ½ a and t2

in this case the positions are y = y₀ = 0 and the accelerations

a = - 1,616m/s2,

we substitute

            0 = 0 + v_{oy} t - ½ a_y t²

            v_{oy}= ½ a_y t

            t = 2v_{oy} / a_y

let's calculate

           t = 2 0.6139 10⁵ / 1.616

           t = 7.597 10⁴s

in this time the particle travels a horizontal distance

           x = v₀ₓ t

           x = 0.8145 10⁵ 7.597 10⁴

           x = 6.19 10⁹ m

the particle falls off the plate

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From rest, a rock is dropped from a garage roof. The roof is 6.0 meters above ground level. The rock will reach the earth at a speed of 10.849 meters per second.

<h3>What is velocity?</h3>

The change of displacement with respect to time is defined as the velocity.  Velocity is a vector quantity.

it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

Given data:

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brainly.com/question/862972

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A rock is dropped from a sea cliff and hits the water 3.2s later. How high is the cliff? How long would it take sound to travel
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Explanation:

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