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aliya0001 [1]
3 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]3 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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STatiana [176]

Answer:

Final velocity = 7.677 m/s

KE before crash = 202300 J

KE after crash = 182,702.62 J

Explanation:

We are given;

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u1 = 17 m/s

u2 = 0 m/s

Using formula for inelastic collision, we have;

m1•u1 + m2•u2 = (m1 + m2)v

Where v is final velocity after collision.

Plugging in the relevant values;

(1400 × 17) + (4700 × 0) = (1400 + 1700)v

23800 = 3100v

v = 23800/3100

v = 7.677 m/s

Kinetic energy before crash = ½ × 1400 × 17² = 202300 J

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3 years ago
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EDIT: He has these backwards, the shortest wavelength is created by Gamma-Rays and the longest is Radiowaves.

<span> Remember- high energy = short wavelength. </span>
6 0
4 years ago
A steel cable has a cross-sectional area 4.49 × 10^-3 m^2 and is kept under a tension of 2.96 × 10^4 N. The density of steel is
Lemur [1.5K]

Answer:

The transverse wave will travel with a speed of 25.5 m/s along the cable.

Explanation:

let T = 2.96×10^4 N be the tension in in the steel cable, ρ  = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.

then, if V is the volume of the cable:

ρ = m/V

m = ρ×V

but V = A×L , where L is the length of the cable.

m = ρ×(A×L)

m/L = ρ×A

then the speed of the wave in the cable is given by:

v = √(T×L/m)

  = √(T/A×ρ)

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  = 25.5 m/s

Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.

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3 years ago
A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the no
padilas [110]

Explanation:

The weight of the car is equal to, W_c=m\times g...........(1)

Where

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g is the acceleration due to gravity

The normal or vertical component of the force is, F_N=mg\ cos\theta

or

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The horizontal component of the force is, F_H=mg\ sin\theta

Taking ratio of equation (1) and (2) as :

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