A diverging lens has a focal length of -30.0 cm. An object is placed 18.0 cm in front of this lens.(a) Calculate the image distance.

(b) Calculate the magnification.

Answers

Answer 1

Answer:

Answer:

A) Calculate the distance

Related Questions

A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, vproton/valpha is:
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Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 * 1014 Hz belong
In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)
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On average, both arms and hands together account for 13 % of a person's mass, while the head is 7.0% and the trunk and legs account for 80 % . We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally. Suppose a 61.0-kg skater is 1.80 m tall, has arms that are each 70.0 cm long (including the hands), and a trunk that can be modeled as being 35.0 cm in diameter. The skater is initially spinning at 70.0 rpm with his arms outstretched.Required:
What will his angular velocity be (in rpm) when he pulls in his arms until they are at his sides parallel to his trunk?

Answers

Final answer:

To find the final angular velocity when the skater pulls in his arms, we use the conservation of angular momentum.

Explanation:

To find the final angular velocity when the skater pulls in his arms, we can make use of the conservation of angular momentum. Initially, the skater's arms are outstretched, and the moment of inertia can be calculated using the parallel axis theorem. After the skater pulls in his arms, we can calculate the new moment of inertia using the same theorem. Equating the initial and final angular momentum values, we can solve for the final angular velocity.

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Final answer:

The problem involves the concept of conservation of angular momentum. The skater's spinning speed will increase when they pull their arms in. For a precise value of the final velocity, a complex calculation taking into account body mass distribution is needed.

Explanation:

This question involves the principle of conservation of angular momentum, which states that the angular momentum of an object remains constant as long as no external torques act on it. The total initial angular momentum of the skater spinning with outstretched arms is equal to his final angular momentum when he pulls his arms in.

Calculating the skater's initial and final angular momentum, you can then solve for his final velocity.

However, note that the calculation needs to take into account the skater's mass distribution. Specifically, we need to consider the percentage distributions for the arms/hands (13%), head (7%) and trunk/legs (80%), and integrate these over the skater's body.

This can result in a significantly complex calculation if done accurately, involving calculus level mathematics. However, using the qualitative knowledge that the skater's spinning speed will increase when they pull their arms in, it's reasonable to estimate, considering the mass distribution, the final velocity will be somewhere near 2 to 3 times the original rpm. But for an exact value, a detailed and complex calculation is needed.

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An ideal step-down transformer is needed to reduce a primary voltage of 120 V to 6.0 V. What must be the ratio of the number of turns in the secondary to the number of turns in the primary

Answers

Answer:

$N_s : N_p = 20 : 1$

Explanation:

From the question we are told that

The primary voltage is $V_p = 120 \ V$

The secondary voltage is $V_s = 6 \ V$

Generally from the transformer equation we have that

$(V_p)/(V_s) = (N_p)/(N_s)$

$(120)/(6) = (N_p)/(N_s)$

=> $(N_p)/(N_s) = 20$

Therefore the ratio of the number of turns in the secondary to the number of turns in the primary is

$N_s : N_p = 20 : 1$

A race car travels on a circular track at an average rate of 125 mi/h. The radius of the track is 0.320 miles. What is the centripetal acceleration of the car? 391 mi/h2 40.0 mi/h2 5,000 mi/h2 48,800 mi/h2

Answers

$a_(centrip)= (v^2)/(R)= (125^2 )/(0.32)=48828 (mi)/(h^2)$
The last one is the closest

48,800 mi/h2 is the right answer

A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 21 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 3.3 s?

Answers

Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad

Two parallel wires I and II that are near each other carry currents i and 3i both in the same direction. Compare the forces that the two wires exert on each other. A. The wires exert equal magnitude attractive forces on each other. B. Wire I exerts a stronger force on wire II than II exerts on I.C. Wire II exerts a stronger force on wire I than I exerts on II. D. The wires exert equal magnitude repulsive forces on each other. E. The wires exert no forces on each other.

Answers

Answer:

A. The wires exert equal magnitude attractive forces on each other.

Explanation:

Magnetic field due to current i on current 2i

B₁ = 10⁻⁷ x 2 i / r where r is distance between the two wires

Force on wire II due to wire I per unit length

= magnetic field x current in wire II

= B₁ x 2 i

= [ 10⁻⁷ x 2 i / r ] x 2i

= 4 x 10⁻⁷ i² / r

Magnetic field due to current 2i on current i

B₂ = 10⁻⁷ x 4 i / r where r is distance between the two wires

Force on wire I due to wire II per unit length

= magnetic field x current in wire I

= B₂ x i

= [ 10⁻⁷ x 4 i / r ] x i

= 4 x 10⁻⁷ i² / r

So final forces on each wire are same .

This force will be attractive in nature . The direction of force can be known from fleming's right hand rule .

A bug flying horizontally at 1.7 m/s collides and sticks to the end of a uniform stick hanging vertically from its other end. After the impact, the stick swings out to a maximum angle of 7.0Â° from the vertical before rotating back. If the mass of the stick is 16 times that of the bug, calculate the length of the stick (in m).