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Ede4ka [16]
2 years ago
11

Where should an object be placed in front of a concave mirror’s principal axis to form an image that is real, inverted, larger t

han the object, and farther from the mirror than the object is?
a. at the focal point on the principal axis
b. at the center of curvature on the principal axis
c. between the center of curvature and focal point on the principal axis
d. between the focal point and the vertex on the principal axis
Physics
2 answers:
malfutka [58]2 years ago
8 0
<em> I believe the answer is letter C.</em>
mart [117]2 years ago
4 0

Answer:

option C

Explanation:

In a concave mirror for the image to be real, inverted  and enlarged the object must be place between the center of curvature and focal point on the principal axis. To obtain an enlarged image object must not in proximity of the mirror, it should be far. The image formed will be on the same side as object( real ). but image will be inverted.

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Calculate Gravitational Potential Energy for an object on Earth with a mass of 2 kg and a height of 7 m.
victus00 [196]

Answer:

137.2J

Explanation:

Ep= mgh

Given,

  • m=2kg
  • h=7m

and we know, g = 9.81 N/kg

Ep= 2 × 9.81 × 7

Ep= 137.2J

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3 years ago
Why was the concern over global cooling replaced with a concern over global warming?
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3 years ago
A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from
ladessa [460]

Answer:

P_1 = 1166.7 Watt

P_2 = 2000 Watt

Explanation:

Average power for the human sprinter is given as

Power = \frac{\Delta E}{\Delta t}

so we have

P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}

P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}

P_1 = 1166.7 Watt

Average power for greyhound is given as

P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}

P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}

P_2 = 2000 Watt

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A_x = 5.0

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6 0
2 years ago
What is the energy range (in joules) of photons of wavelength 410 nm to 750 nm ? Express your answers using two significant figu
andreyandreev [35.5K]

Answer:

4.9 x 10^-19 J, 2.7 x 10^-19 J

Explanation:

first wavelength, λ1 = 410 nm = 410 x 10^-9 m

Second wavelength, λ2 = 750 nm = 750 x 10^-9 m

The relation between the energy and the wavelength is given by

E = h c / λ

Where, h is the Plank's constant and c be the velocity of light.

h = 6.63 x 10^-34 Js

c = 3 x 10^8 m/s

So, energy correspond to first wavelength

E1 = (6.63 x 10^-34 x 3 x 10^8) / (410 x 10^-9) = 4.85 x 10^-19 J

E1 = 4.9 x 10^-19 J

So, energy correspond to second wavelength

E2 = (6.63 x 10^-34 x 3 x 10^8) / (750 x 10^-9) = 2.652 x 10^-19 J

E2 = 2.7 x 10^-19 J

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