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2 years ago

Can anyone please tell me the properties of image formed in a spherical mirror?

Physics

2 answers:

notsponge [240]2 years ago

6 0

Answer:

The two rays, CY and DM are diverging rays and when extended behind the mirror, they appear to intersect each other at point M'. Therefore, the properties of the images formed here are formed behind the mirror, between the pole and principal focus (f), the images are diminished and are virtual and erect.

Explanation:

<h2>Spherical Mirrors</h2>

There are two kinds of spherical mirrors, concave and convex.

The focal point (F) of a concave mirror is the point at which a parallel beam of light is "focussed" after reflection in the mirror. ...

The focal length (f) and radius of curvature (R) are defined in the diagram at the right.

<h3>hope it helps and thanks for following </h3><h2>please give brainliest </h2>

Llana [10]2 years ago

4 0

Answer:

This questions answers are on the baisis of where the object is placed and see the atachments to know more

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A hockey puck slides 55.0 m along the length of the rink in just 1.25 s. The slight friction between the puck and the ice provid

topjm [15]

Answer:

44.8 m/s

Explanation:

Use the Initial Speed Formula:

InS = 2(d/t) - Final Speed

InS = 2(55/1,25) - 43.2

InS = 2.44 - 43,2

InS = 88 - 43,2

InS = 44.8 m/s

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2 years ago

How are motors and generators different?

saw5 [17]

A generator converts mechanical energy into electrical energy, while a motor does the opposite - it converts electrical energy into mechanical energy

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3 years ago

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece

Brilliant_brown [7]

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

$v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m$

d = 10.01 cm

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2 years ago

A river flows with a speed of 0.600 m/s. A student first swims upriver 0.500 km, then turns around and returns to his starting p

DerKrebs [107]

Answer:

a) 1111.0 seconds

b) 833.3 s

c) Because of proportions

Explanation:

a) Total time of round trip is the sum of time upriver and time downriver

$t_{total}=t_{up}+t_{down}$

Time upriver is calculated with the net speed of student and 0.500 km:

$t_{up}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=v_{relative to river}+v_{river}=-1.2+0.6=-0.6 m/s\\t_{up}=\frac{500 m}{0.6 m/s}=833.3 s$

(Becareful with units 0.5 km= 500m) Similarly of downriver:

$t_{down}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=1.2+0.6=1.8 m/s\\t_{down}=\frac{500 m}{1.8 m/s}=277.7 s$

So the sum is:

$t_{total}=1111.0s$

b) Still water does not affect student speed, so total time would be simply:

$t_{total}=\frac{1000 m}{1.2 m/s}=833.3 s$

c) For the upriver trip, student moved half the distance in half speed of the calculation in b), so it kept the same ratio and therefore, same time. So the aditional time is actually the downriver.

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3 years ago

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Gala2k [10]

Electrical energy.................

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3 years ago