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marusya05 [52]
2 years ago
6

Please i need detailed explanation​

Physics
1 answer:
zubka84 [21]2 years ago
4 0

Answer:

2Micro Farahds

Explanation:

Its in the picture.

I Hope it helps.

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If Margo pushes a box 25 meters, using a force of 45 Newtons, and it takes her 15 seconds to do so, how much power is she using?
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She is using 75 watt power.
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A plate in a parallel-plate capacitor has an area of 0.03 m2 and is 0.5 × 10–3 m from the other plate. The space between the pla
aivan3 [116]

Answer:

4 x 10^-9F

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3 years ago
true or false:acceleration toward the center of a curved or circular path is called gravitational acceleration.
nalin [4]
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5 0
2 years ago
From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the
Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

           x = Vox t

           t = x / vox

           t = 7.1 / 340

           t = 2.09 10-2 s

In this same time the height of the window fell

           Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

           Voy = (Y + ½ g t²) / t

           Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

            Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

             Vy = Voy - gt₂

             Vy = 0 -g t₂

             t₂ = Vy / g

             t₂ = 27.7 / 9.8

             t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

            X = Vox t₂

            D = 340 2.83

            D = 962.2 m

           

            Y = Voy₂– ½ g t₂²

            Y = 0 - ½ g t2

            H = Y = - ½ 9.8 2.83 2

            H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

6 0
3 years ago
Differentiate between atmospheric pressure and pressure.​
Dovator [93]

Answer: atmospheric is air by the earth and pressure is just someone or something doing it

Explanation:

8 0
2 years ago
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