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aliya0001 [1]
2 years ago
7

Consider a concave mirror that has a focal length f. In terms of f, determine the object distances that will produce a magnifica

tion of
A. -2
B. -3
C. -4
Physics
1 answer:
Aleks04 [339]2 years ago
3 0

We have that the magnification of each focal length is given respectively as

A) has u=3\frac{f}{2}

B) has u=4\frac{f}{3}

C) has  u=5\frac{f}{4}

From the question we are told that:

Focal Length F

Generally, the equation for Magnification is mathematically given by

M=\frac{-v}{u}

Therefore

v=2u

For A

M=-2

Therefore

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{2u}

Therefore

u=3\frac{f}{2}

For B

M=-3

Therefore

v=3u

Where

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{3u}

Therefore

u=4\frac{f}{3}

For C

M=-4

Therefore

v=4u

Therefore

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{4u}

Therefore

u=5\frac{f}{4}

Conclusion

From the calculations above we can rightly say that the magnifications of the values above are

A has u=3\frac{f}{2}

B has u=4\frac{f}{3}

C has  u=5\frac{f}{4}

For more information on this visit

brainly.com/question/14468351

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A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the b
mars1129 [50]

Answer:

ω2  =  216.47 rad/s

Explanation:

given data

radius r1 =  460 mm

radius r2 = 46 mm

ω =  32k rad/s

solution

we know here that power generated by roller that  is

power = T. ω    ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1)  ÷ 2 = (  F × r2 × ω2 )  ÷  2    ................2

so

ω2  =  \frac{460\times 32}{34\times 2}

ω2  =  216.47

8 0
3 years ago
A circular loop of wire of area 10 cm^2 carries a current of 25 A. At a particular instant, the loop lies in the xy-plane and is
S_A_V [24]

The magnetic dipole moment of the current loop is  0.025 Am².

The magnetic torque on the loop is 2.5 x 10⁻⁴  Nm.

<h3>What is magnetic dipole moment?</h3>

The magnetic dipole moment of an object, is the measure of the object's tendency to align with a magnetic field.

Mathematically, magnetic dipole moment is given as;

μ = NIA

where;

  • N is number of turns of the loop
  • A is the area of the loop
  • I is the current flowing in the loop

μ = (1) x (25 A) x (0.001 m²)

μ = 0.025 Am²

The magnetic torque on the loop is calculated as follows;

τ = μB

where;

  • B is magnetic field strength

B = √(0.002² + 0.006² + 0.008²)

B = 0.01 T

τ = μB

τ =  0.025 Am² x 0.01 T

τ = 2.5 x 10⁻⁴  Nm

Thus, the magnetic dipole moment of the current loop is determined from the current and area of the loop while the magnetic torque on the loop is determined from the magnetic dipole moment.

Learn more about magnetic dipole moment here: brainly.com/question/13068184

#SPJ1

8 0
1 year ago
A single loop of wire with an area of 0.0900 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic
erica [24]

Answer:

(a) 0.0171 V

Explanation:

A = 0.09 m^2, dB/dt = 0.190 T/s

(a) According to the law of electromagntic induction

e = dФ / dt

e = A dB / dt

e = 0.09 x 0.190 = 0.0171 V

(b)

as we know

i = e / R

we can find induced current by dividing induced emf by resistance

5 0
3 years ago
The quadriceps muscles pull on the patella simultaneously. Below are the forces from each
Nostrana [21]

Based on the calculation of the resultant of vector forces:

  1. the resultant force due to the quadriceps is 1795 N
  2. the resultant force due to the quadriceps is 1975 N. Training and strengthening the vastus medialis results in a greater force of muscle contraction.

<h3>What is the resultant force due to the quadriceps?</h3>

The resultant of more than two vector forces is given by:

  • F = √Fₓ² + Fₙ²

where:

  • Fₓ is the sum of the horizontal components of the forces
  • Fₙ is the sum of the vertical components of the forces
  • Fx = F₁cosθ + F₂cosθ + F₃cosθ + F₄cosθ
  • Fₙ = F₁sinθ + F₂sinθ + F₃sinθ + F₄sinθ
  • F₁ = 680N, θ = 90 = 30 = 120°
  • F₂ = 220 N, θ = 90 + 16 = 106°
  • F₃ = 600 N, θ = 90 + 15 = 105°
  • F₄ = 480 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 480 * cos 55

Fx = -280.6 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 480 * sin 55

Fₙ = 1773.1 N

then:

F = √(-280.6)² + ( 1773.1)²

F = 1795.16 N

F ≈ 1795 N

Therefore, the resultant force due to the quadriceps is 1795 N

<h3>What would happen if the vastus medialis was trained and strengthened to contract with 720N of force?</h3>

From the new information provided:

  • F₁ = 680N, θ = 90 = 30 = 120°
  • F₂ = 220 N, θ = 90 + 16 = 106°
  • F₃ = 600 N, θ = 90 + 15 = 105°
  • F₄ = 720 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 720 * cos 55

Fx = -142.95 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 720 * sin 55

Fₙ = 1969.72 N

then:

F = √(-142.95)² + ( 1969.72)²

F = 1974.9 N

F ≈ 1975 N

Therefore, the resultant force due to the quadriceps is 1975 N.

Training and strengthening the vastus medialis results in a greater force of muscle contraction.

Learn more about resultant of forces at: brainly.com/question/25239010

3 0
2 years ago
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