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elixir [45]
3 years ago
12

A pressure antinode in a sound wave is a region of high pressure, while a pressure node is a region of low pressure.

Physics
1 answer:
maks197457 [2]3 years ago
3 0
A pressure antinode in a sound wave is not a region of high pressure, while a pressure node is not a region of low pressure.

The answer is false

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Two particles are fixed to an x axis: particle 1 of charge −8.00 ✕ 10⁻⁷ C at x = 6.00 cm, and particle 2 of charge +8.00 ✕ 10⁻⁷
victus00 [196]

Answer:

0 N/C

Explanation:

Parameters given:

q_1 = -8.00 * 10^{-7} C

x_1 = 6.00 cm

q_2 = +8.00 * 10^{-7} C

x_2 = 21 cm

The distance between q_1 and q_2 is

21 - 6 = 15cm

Electric field is given as

E = \frac{kq}{r^2}

r = 15/2 = 7.5cm = 0.075m

The electric field at their midpoint due to q_1 is:

E = \frac{9 * 10^9 * -8.0 * 10^{-7}}{0.075^2}

E_1 = -1.28 * 10^6 N/C

The electric field at the midpoint due to q_2 is:

E = \frac{9 * 10^9 * 8.0 * 10^{-7}}{0.075^2}

E_2 = 1.28 * 10^6 N/C

The net electric field will be:

E = E_1 + E_2

E = -1.28 * 10^6 + 1.28 * 10^6

E = 0 N/C

7 0
4 years ago
A ball is thrown at an angle of 45° to the ground. if the ball lands 87 m away, what was the initial speed of the ball? (round
Troyanec [42]
Just try your best best friend everyone
6 0
4 years ago
A circuit has a current of 1.2 A. If the voltage decreases to one-third of its original amount while the resistance remains cons
lapo4ka [179]
Consider the formula ...

Current  =  (voltage) / (resistance) .

If the resistance doesn't change, then the current is
directly proportional to the voltage.

If the voltage decreases to 1/3 of what it used to be,
then the current does the same thing.

1.2 A  ---->  0.4 A
6 0
3 years ago
What isthe correct 1/4=20%
anzhelika [568]

Answer:

25

Explanation:

5 0
3 years ago
21.-Una esquiadora olímpica que baja a 25m/s por una pendiente a 20o encuentra una región de nieve húmeda de coeficiente de fric
jeyben [28]

Answer:

y = 12.82 m

Explanation:

We can solve this exercise using the energy work theorem

          W = ΔEm

friction force work is

          W = fr . s = fr s cos θ

the friction force opposes the movement, therefore the angle is 180º

           W = - fr s

we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular

           N -Wy = 0

           N = mg cos θ

the friction force remains

            fr = μ N

            fr = μ mg cos θ

             

work gives

           W = - μ mg s cos θ

initial energy

           Em₀ = ½ m v²

the final energy is zero, because it stops

we substitute

          - μ m g s cos θ = 0 - ½ m v²

          s = ½ v² / (μ g cos θ)

         

let's calculate

          s = ½ 20² / (0.55 9.8 cos 20)

          s = 39.49 m

this is the distance it travels along the plane, to find the vertical distance let's use trigonometry

            sin 20 = y / s

           y = s sin 20

           y = 37.49 sin 20

           y = 12.82 m

8 0
4 years ago
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