Ask question

Ask question

All categories

3 years ago

7

A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?

1 answer:

Makovka662 [10]3 years ago

5 0

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length $f=\frac{R}{2}=\frac{12}{2}=6cm$

Now for mirror we know that $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

So $\frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}$

$16.66-0.333=\frac{1}{v}$

v = 0.750 m

Now magnification of the mirror is $m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025$

You might be interested in

A spacecraft of mass 1500 kg orbits the earth at an altitude of approximately 450 km above the surface of the earth. Assuming a

PSYCHO15rus [73]

Answer:

1.28 x 10^4 N

Explanation:

m = 1500 kg, h = 450 km, radius of earth, R = 6400 km

Let the acceleration due to gravity at this height is g'

g' / g = {R / (R + h)}^2

g' / g = {6400 / (6850)}^2

g' = 8.55 m/s^2

The force between the spacecraft and teh earth is teh weight of teh spacecraft

W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N

8 0

3 years ago

Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 44.3 ◦ . The velo

Goshia [24]

Answer:

So airplane will be 1324.9453 m apart after 2.9 hour

Explanation:

So if we draw the vectors of a 2d graph we see that the difference in angles is = 83 - 44.3 = $83-44.3=38.7^{\circ}$

Distance traveled by first plane = 730×2.9 = 2117 m

And distance traveled by second plane = 590×2.9 = 1711 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.

Using the law of cosine, $d^2$ representing the distance between the planes, we see that:

$d^2=2117^2 + 1711^2 -2\times (2117)\times (1711)cos(38.7)=1755480.2482$

d = 1324.9453 m

4 0

3 years ago

Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t

Musya8 [376]

Answer:

$n = 1.266\times 10^{12}$

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is $\theta = 13 degree$

thickness of string 300 mm = 0.3 m

$sin\theta =\frac{2}{0.3 m}$

r =0.3 sin 13 = 0.067 m

$Fe = \frac{ kq_1 q-2}{d^2}$

$Fe = \frac{kq^2}{(2r)^2} = mg tan\theta$

$q^2 = mg tan\theta \frac{(2r)^2}{k}$

$= 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}$

$q^2 = 4.10\times 10^{-14}$

$q = 2.026 \times 10^{-7} C$

q = ne

$n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}$

$n = 1.266\times 10^{12}$

3 0

3 years ago

A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

Tema [17]

Answer: $410.90\times 10^{-6} V$

Explanation:

Given

$A=5.5 mm^2$

$n=919 turns/cm\approx 85400 turns/m$

$\omega =226 rad/s$

According to the Faraday law of induction, induced emf is given by

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

Magnetic field $\phi _B=B\cdot A$

$B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )$

$B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)$

$B=0.3305\sin (226t)$

$\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}$

$\phi _{B}=1.818\times 10^{-6}\sin (226t)$

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

$E=-1.818\times 10^{-6}\times 226\cos (226t)$

$E=-410.90\times 10^{-6}\cos (226t)$

Amplitude of EMF induced $=410.90\times 10^{-6} V$

8 0

3 years ago

HELP PLEASE <br> Find the net force necessary for a 15 kg object to accelerate at 20 m/s/s.

tensa zangetsu [6.8K]

Answer:

<h3>The answer is 300 N</h3>

Explanation:

The force acting on an object given the mass and acceleration we use the formula

<h3>force = mass × acceleration</h3>

We have

force = 15 × 20

We have the final answer as

<h3>300 N</h3>

Hope this helps you

3 0

3 years ago