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Pani-rosa [81]
2 years ago
10

The focal length of a diverging lens is negative. If f = −26 cm or a particular diverging lens, where will the image be formed o

f an object located 63 cm to the left of the lens on the optical axis?
.... cm to the left of the lens

....What is the magnification of the image?
Physics
1 answer:
Arada [10]2 years ago
7 0

The distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

<h3>Location of the image</h3>

The image distance formed by the lens is calculated as follows;

1/v + 1/u = -1/f

where;

  • v is the image distance
  • u is the object distance
  • f is the focal length of the lens

1/v = -1/f - 1/u

1/v = -(-1/26) - 1/63

1/v = 1/26 - 1/63

1/v = 0.022588

v = 1/0.022588

v = 44.27 cm

<h3>What is magnification of lens?</h3>

The magnification of a lens is defined as the ratio of the height of an image to the height of an object.

<h3>Magnification of the image formed</h3>

The magnification of the image is calculated as follows;

Magnification = image distance/object distance

M = 44.27/63

M = 0.7

Thus, the distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

Learn more about magnification here: brainly.com/question/1599771

#SPJ1

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Vilka [71]

See the attached picture:

8 0
3 years ago
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You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,
lbvjy [14]

Answer:

The value is T_c  =  12 .1 ^oC

Explanation:

From the question we are told that

The mass of the ice cube is m_i  =  7.50 *10^{-3} \  kg

The temperature of the ice cube is T_i = 0^o C

The mass of the copper cube is m_c  =  0.540 \  kg

The final temperature of both substance is T_f  =  0^oC

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

Q =  m_c  *  c_c *  [T_c  -  T_f ]

The specific heat of copper is c_c  = 385J/kg \cdot  ^oC

Generally the heat gained by the ice cube is mathematically represented as

Q_1 =  m_i * L

Here L is the latent heat of fusion of the ice with value L  =  3.34 * 10^{5} J/kg

So

Q_1 =  7.50 *10^{-3} * 3.34 * 10^{5}

=> Q_1 =  2505 \ J

So

2505  =  0.540  *  385 *  [T_c  - 0 ]

=>    T_c  =  12 .1 ^oC

4 0
3 years ago
What is the answer to this question?
dem82 [27]
Acceleration is a change in *speed* over time. In this case, the speed of the car increased by 90 km/hr in 6 s, giving it a rate of 90 km/hr/6s, or 15 km/hr/s. We’re asked for the acceleration in m/s^2, though, so we’ll need to do a few conversions to get our units straight.

There are 1000 m in 1 km, 60 min, or 60 * 60 = 3600 s in 1 hr, so we can change our rate to:

(15 x 1000)m/3600s/s, or (15 x 1000)m/3600 s^2

We can reduce this to:

(15 x 10)m/36 s^2 = 150 m/36 s^2

Which, dividing numerator and denominator by 36, gets us a final answer of roughly 4.17 m/s^2
3 0
3 years ago
An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel s
Goshia [24]

Answer:

1) θ = 0.00118 rad, 2)  θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4)  I / Io = 0.216

Explanation:

In the double-slit interference phenomenon it is explained for constructive interference by the equation

          d sin θ = m λ

1) the first order maximum occurs for m = 1

           sin θ = λ  / d

           θ = sin⁻¹ λ  / d

let's reduce the magnitudes to the SI system

           λ  = 520 nm = 520 10⁻⁹  θ = 0.00118 radm

           d = 0.440 mm = 0.440 10⁻³ m ³

let's calculate

           θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)

            θ = sin⁻¹ (1.18 10⁻³)

            θ = 0.00118 rad

2) the second order maximum occurs for m = 2

            θ = sin⁻¹ (m λ  / d)

            θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)

            θ = 0.00236 rad

3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains

          I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )

where the function sinc = sin x / x

and b is the width of the slits

we caption the values

             x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)

             x = 2.21

            I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²

remember angles are in radians

            I / I₀ = cos² (3.0945) [0.363] 2

            I / I₀ = 0.9978 0.1318

            I / I₀ = 0.1738

4) the maximum second intensity is

            I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)

            x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)

            x = 4.41

            I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²

            I / Io = cos² 6.273    0.216

            I / Io = 0.216

.

7 0
3 years ago
What is the potential energy of a 2 kg ball 15 m in the air?
oee [108]

Answer:

<h2>300 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 2 × 10 × 15

We have the final answer as

<h3>300 J</h3>

Hope this helps you

7 0
3 years ago
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