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tangare [24]
3 years ago
6

If you were standing at the center of curvature in front of a concave mirror, what image would be projected?

Physics
2 answers:
matrenka [14]3 years ago
6 0
"<span>The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are" is the type of image among the choices given in the question that would be projected. The correct option among all the options that are given in the question is the first option. I hope it helps you.</span>
Bas_tet [7]3 years ago
5 0

Answer:

The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

Explanation:

As we know that the object position is at center of curvature of the mirror

So here the distance is given as

d = R = 2f

now by mirror formula of relation between distance of image and distance of object

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + \frac{1}{2f} = \frac{1}{f}

d_i = 2f

so image will form at the position of object itself

Now we will say that the magnification of the object will be

M = -\frac{d_i}{d_o} = -1

so image will be upside down and same as that the height of object

so correct answer would be

The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

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maria [59]

Answer:

12 ml

Explanation:

The initial volume in the cylinder is 20 ml

 adding the rock adds volume to the cylinder

       the new volume is 32 ml .....the increase in volume is the volume of the rock :   32 - 20 = 12 ml    volume of rock

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2 years ago
What are the names of the 2 charged particles in an atom and what are their charges
Sholpan [36]

Answer:

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Explanation:

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  • The physical and chemical characteristics of an atom are determined by the number of these particle composition.
  • The protons present in the center of an atom called the nucleus is positively charged.
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3 0
3 years ago
(1 pt) A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 35 meters above the ground. This takes 14
Elenna [48]

Answer:

w = 5832.372 Joules

Explanation:

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The water was pulled up to a height of 35 meters, i.e. h = 35 m

It takes 14 minutes to pull up the water through the height, 35 m

speed = distance/ time = 35/14 = 2.5 m/min

The bucket's height, y = speed * time = 2.5t meters

6 kg of water drips out of the bucket throughout the 14 minutes

The rate at which the water drips drips out = (6/14) = 0.4286 kg/min

Mass of water that drips out in time, t = 0.4286t kg

The mass of water remaining = (20 - 0.4286t) kg

Change in Workdone, Δw = mgΔy

Δy = 2.5 Δt

Δw = mg *  2.5 Δt

dw =  (20 - 0.4286t)g2.5 dt

integrating both sides

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w = \int\limits^a_b {(50g-1.07gt)} \, dx where b = 0, a = 14

w = 50gt - 1.07g(t²)/2      g = 9.8 m/s²

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w = 6860 - 1027.628

w = 5832.372 Joules

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