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yulyashka [42]
2 years ago
9

An erect object is placed on the central axis of a thin lens, further from the lens than the magnitude of its focal length. The

magnification is +0.4. This means:
A) the image is real and erect and the lens is a converging lens
B) the image is real and inverted and the lens is a converging lens
C) the image is virtual and erect, and the lens is a diverging lens
D) the image is virtual and erect, and the lens is a converging lens
E) the image is virtual and inverted and the lens is a diverging lens
Physics
1 answer:
Zolol [24]2 years ago
4 0

Answer:

the image is virtual and erect and the lens divergent; therefore the correct answer is C

Explanation:

In a thin lens the magnification given by

      m = h '/ h = - q / p

where h ’is the height of the image, h is the height of the object, q is the distance to the image and p is the distance to the object.

It indicates that the object is straight and is placed at a distance p> f

analyze the situation tells us that the magnification is positive so the distance to the image must be negative, that is, that the image is on the same side as the object.

Consequently the lens must be divergent

The magnification value is

          0.4 = h ’/ h

          h ’= 0.4 h

therefore the erect images

therefore the image is virtual and erect and the lens divergent; therefore the correct answer is C

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A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f
konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m

(c) the maximum speed attained by the object during its motion

E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s

8 0
2 years ago
A runner runs 4875 ft in 6.85 minutes. what is the runners average speed in miles per hour?
yanalaym [24]

The average speed can be easily calculated by taking the ratio of distance and time. That is:

average speed = distance / time

 

so calculating:

average speed = 4875 ft / 6.85 minutes

<span>average speed = 711.68 ft / min</span>

8 0
2 years ago
Read 2 more answers
An electric drill transfers 200 J of energy into a useful kinetic energy store. It also transfers 44 J of energy by sound and 48
kramer

Answer:

Er = 108 [J]

Explanation:

To solve this problem we must understand that the total energy is 200 [J]. Of this energy 44 [J] are lost in sound and 48 [J] are lost in heat. In such a way that these energy values must be subtracted from the total of the kinetic energy.

200 - 44 - 48 = Er

Where:

Er = remaining energy [J]

Er = 108 [J]

3 0
1 year ago
1. A car is 140 kg, and drove East 13.5 m/s. Car B is 157 kg and drove West at 10.9 m/s.
vaieri [72.5K]
The velocity would switch on the cars
4 0
2 years ago
An electron is at the origin. (a) Calculate the electric potential VA at point A, x 5 0.250 cm. (b) Calculate the electric poten
saw5 [17]

Answer:

a)  V_a = -5.7536 10⁺⁷ V , b) Vb = -1.92 10⁻⁷ V  c) the sign of the potential change

Explanation:

The electrical potential for a point charge

     V = k q / r

Where k is the Coulomb constant that you are worth 8.99 10⁹ N m² / C²

a) potential At point x = 0.250 cm = 0.250 10-2m

    V_a =  -8.99 10⁹ 1.6 10⁻¹⁹ /0.250 10⁻²

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b) point x = 0.750 cm = 0.750 10-2

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    Vb = -1.92 10⁻⁷ V

potemcial difference

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    V_ba = (-5.7536 + 1.92) 10⁻⁷

    V_ba = -3.83 10⁻⁷ V

c) To know what would happen to a particle, let's use the relationship between the potential and the electric field

     ΔV = E d

The force on the particle is

     F = q₀ E

     F = q₀ ΔV / d

We see that the force on the particle depends on the sign of the burden of proof. Now the burden of proof is negative to pass between the two points you have to reverse the sign of the potential, bone that the value should be reversed

          V_ba = 0.83 10⁻⁷ V

5 0
2 years ago
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