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3 years ago

5

A(n) ____________ image cannot be projected and forms where light rays appear to originate.

2 answers:

IrinaVladis [17]3 years ago

6 0

A virtual image is the answer.

Sedbober [7]3 years ago

6 0

<h3><u>Answer;</u></h3>

Virtual image

A<em><u> virtual image</u></em> cannot be projected and forms where light rays appear to originate.

<h3><u>Explanation</u>;</h3>

<em><u>A virtual image is an image that can not be formed on a screen. This type of an image is formed when the rays of light appear to meet at a point after reflection or refraction. Virtual images are always erect or upright.</u></em>
<em><u>A real image on the other hand, is an image that can be formed on a screen. Real image is formed when the rays of light meet at some point after reflection or refraction. Real images are always inverted.</u></em>

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Electromagnetic radiation from a 8.25 mW laser is concentrated on a 1.23 mm2 area. Suppose a 1.12 nC static charge is in the bea

mylen [45]

Answer:

The maximum magnetic force is 2.637 x 10⁻¹² N

Explanation:

Given;

Power, P = 8.25 m W = 8.25 x 10⁻³ W

charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C

speed of the charge, v = 314 m/s

area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²

The intensity of the radiation is calculated as;

$I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2$

The maximum magnetic field is calculated using the following intensity formula;

$I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T$

The maximum magnetic force is calculated as;

F₀ = qvB₀

F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)

F₀ = 2.637 x 10⁻¹² N

5 0

3 years ago

List at least 3 units of speed circle the unit of speed which is the SI unit

Stells [14]

Answer: SI Unit is meters per second(m/s)

Explanation:

The three units of speed are:

1. meters per second(m/s)

2. kilometers per hour(km/hr)

3. miles per hour(mph)

SI Unit is meters per second(m/s)

6 0

3 years ago

5. How many kilowatt-hours of energy would be used by a 40 W bulb that runs for 10 hours every

Nady [450]

Answer:

E = 0.4 kWh

Explanation:

Given that,

The power of a bulb, P = 40 W

Time, t = 10 hours

We need to find the energy used by the bulb. Let it is equal to E. So,

$E=P\times t\\\\E=40\times 10\\\\E=400\ W-h\\\\or\\\\E=0.4\ kWh$

So, 0.4 kWh of energy would be needed.

6 0

3 years ago

A small metal bead, labeled A has a charge of 25 nC. It is touched to metal bead B, initially neutral, so that the two beads sha

Iteru [2.4K]

Answer:

15 nC and 10 nC

Explanation:

qA + qB = 25 nC = 25 x 10^-9 C

F = 5.4 x 10^-4 N

d = 5 cm = 0.05 m

Use the Coulomb's law

$F=\frac{Kq_{A}q_{B}}{d^{2}}$

By substituting the values, we get

$5.4 \times 10^{-4}=\frac{9 \times 10^{9}q_{A}q_{B}}{0.05^{2}}$

qA x qB = 1.5 x 10^-16 C

So, $q_{A}\left ( 25 \times 10^{-9}-q_{A} \right )=1.5 \times 10^{-16}$

$q_{A}^{2}-25 \times 10^{-9}q_{A}+1.5 \times 10^{-16}=0$

$q_{A}=\frac{25\times 10^{-9} \pm \sqrt{6.25\times 10^{-16}-6 \times10^{-16}}}{2}$

$q_{A}=\frac {25\times 10^{-9} \pm 5 \times 10^{-9}}}{2}$

qA = 15nC or 10nC

So, qB = 10 nC or 15 nC

8 0

4 years ago

Consider an aircraft with a finite wing area of 60.0 m2 and an aspect ratio of 8. Assume the wing is composed of a NACA 65-210 a

kaheart [24]

Answer:

CL = 0.57

CD = 0.027

Explanation:

Thinking process:

Let the parameters be:

wing area = 21.5 m²

aspect ratio = 5

span efficiency factor = 0.9

CD₀ = 0.004

Angle AOA = 6°

Therefore,

$CD = CD_{0} + \frac{(CL)^{2} }{ITEAR}$

For the NACA 65210

α = 9° ; CL = 1.05

α = -1.5 ; Cl = 0

Therefore, $a_{0} = \frac{1.05- 0}{9-(-1.5)} \\ = 0.11$

Lift slope for finite wing is given by:

$a = \frac{a_{0} }{1+\frac{a_{0} }{ITEAR} } \\ = \frac{0.11}{1+\frac{57.3(0.11)}{\pi }(0.9)(5) }\\ = 0.076 deg$

at α = 6°, CL is given by:

a ( $\alpha -\alpha _{L=0}) = 0.076(6-(-1.5)\\ CL= 0.57$

CD = $0.004 + \frac{(0.57)^{2} }{\pi*0.9*6 }$

= 0.027

6 0

4 years ago