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Ray Of Light [21]
3 years ago
12

Under which conditions do both convex and concave mirrors form virtual images? State whether or not each can form a real image.I

f the mirror can form a real image, describe how this occurs.
Physics
2 answers:
natali 33 [55]3 years ago
8 0
Concave mirror​s can make a real image.when the object is away from the mirror it is called real image and when the object is closer to the mirror it is called virtual mirror.
leva [86]3 years ago
6 0

Answer:

In case of concave mirror the real image will form when object is placed at a distance which is more than the focal length of the mirror.

So here if more distance from the mirror then we will have more chance to get real image.

Now if we wish to have virtual image using concave mirror then we can say that object must have to place at distance less than the focal length of the mirror.

So if object is placed closer to the mirror then the image must be virtual and erect.

Now in case of convex mirror we will say that image formed will be virtual image in all cases because the focal length of the convex mirror is always positive so it will always form virtual image.

So here concave mirror if distance of object is less than the focal length then it will form virtual image and if the distance of object is more than focal length then it will form real image. While is convex mirror it will always form virtual image.

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A person wants to make a metronome for music practice. He uses a 35-g object attached to a spring to serve as the time standard.
alukav5142 [94]

To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Here,

k = Spring constant

m = Mass

Our values are given as,

m = 35g = 35*10^{-3}kg

f = 1 Hz

Rearranging to find the spring constant we have that,

k = (2\pi f \sqrt{m})^2

k = 4\pi^2 f^2 m

k = (4) (\pi)^2 (1) (35*10^{-3})

k = 1.38N/m

Therefore the spring constant is 1.38N/m

7 0
3 years ago
A person hits a tennis ball with a mass of 0.058 kg against a wall.
horrorfan [7]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

F=\dfrac{m(v-u)}{t}

F=\dfrac{0.058\times (11-(-11))}{2.1}

F = 0.607 N

(b) Area of wall, A=3\ m^2

Let P is the average pressure on that area. It is given by :

P=\dfrac{F}{A}

P=\dfrac{0.607\ N}{3\ m^2}

P = 0.202 Pa

Hence, this is the required solution.

8 0
3 years ago
A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.8 m/s rel
Svetach [21]

I'm assuming the question is what is the robin's speed relative to to the ground...

Create an equation that describes its relative motion.

rVg = rVa + aVg


Substitute values.

rVg = 12 m/s [N] + 6.8 m/s [E]


Use vector addition.

| rVg | = √ | rVa |² + | aVg |²

| rVg | = √ 144 m²/s² + 46.24 m²/s²

| rVg | = √ 19<u>0</u>.24 m²/s²

| rVg | = 1<u>3</u>.78 m/s


Find direction.

tanФ = aVg / rVa

tanФ = 6.8 m/s / 12 m/s

Ф = 29°


Therefore, the velocity of the robin relative to the ground is 14 m/s [N29°E]

4 0
3 years ago
Hooke’s law states that the distance that a spring is stretched by hanging object varies directly as the mass of the object. If
trasher [3.6K]

Answer:

d_{2} = 33.33 cm

Explanation:

Given:

When mass, m_{1} =21 kg

          distance travelled is  d_{1}  = 140 cm

When mass, m_{2} =5 kg

         distance travelled is  d_{2}  = ?

Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.

Therefore according to the question,

\frac{d_{1}}{m_{1}}=\frac{d_{2}}{m_{2}}

\frac{140}{21}=\frac{d_{2}}{5}

d_{2} = 33.33 cm

Distance travelled is 33.33 cm when mass is 5 kg.

8 0
2 years ago
How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 4.00
klio [65]

Explanation:

Let us assume that the separation of plate be equal to d and the area of plates is 9 \times 10^{-4} m^{2}. As the capacitance of capacitor is given as follows.

            C = \frac{\epsilon_{o}A}{d}

It is known that the dielectric strength of air is as follows.

               E = 3 \times 10^{6} V/m

Expression for maximum potential difference is that the capacitor can with stand is as follows.

                       dV = E × d

And, maximum charge that can be placed on the capacitor is as follows.

               Q = CV

                   = \frac{\epsilon_{o} A}{d} \times E \times d

                   = \epsilon_{o}AE

                   = 8.85 \times 10^{-12} \times 3 \times 10^{6} \times 4 \times 10^{-4}

                   = 1.062 \times 10^{-8} C

or,                = 10.62 nC

Thus, we can conclude that charge on capacitor is 10.62 nC.

5 0
3 years ago
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