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2 years ago

What is the greatest distance an image can be located behind a convex spherical mirror?

Physics

1 answer:

Afina-wow [57]2 years ago

6 0

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

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

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

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

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

so here maximum value of image distance is equal to focal length of the mirror

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A motor is the devise

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Let say for every 5 s of time interval the speed will remain constant

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so here we will have

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2 years ago

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in a circuit, a 10 resistor is connected in series to a parallel group containing a 60 resistor and a 5 resistor. what is the to

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The 60 and the 5 in parallel have an equivalent resistance of 4.615 ohms. (rounded). The 10 in series makes it 14.615...

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A 3.0 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determi

krok68 [10]

Answer:

a) k = 2231.40 N/m

b) v = 0.491 m/s

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

(m)×(v^2) = (k)×(x^2)

a) (m)×(v^2) = (k)×(x^2)

k = [(m)×(v^2)]/(x^2)

k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)

k = 2231.40 N/m

Therefore, the force spring constant is 2231.40 N/m

b) (m)×(v^2) = (k)×(x^2)

v^2 = [(k)(x^2)]/m

v = \sqrt{ [(k)(x^2)]/m}

v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}

= 0.491 m/s

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2 years ago

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Answer:

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3 years ago