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Maslowich
2 years ago
14

When the distance between a point source of light and a light meter is reduced from 6.0m to 2.0 m, the intensity of illumination

at the meter will be the original value multiplied by _____.
Physics
1 answer:
Sergeeva-Olga [200]2 years ago
7 0

Answer:

Explanation:

Let the point source have power P .

At distance r , the intensity I

I = P / 4πr² . If intensity at 6 m and 2 m be I₁ and I₂

I₁ = P / 4π x 6²

I₂ =  P / 4π x 2²

I₁ / I₂ = 2² / 6²

= 1 / 9

I₂ = 9 I₁

Intensity will be 9 times that at 6 m .

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(PLEASE HELP) Identify the following as electromagnetic (E) or mechanical (M) waves.
fomenos
Sound waves (m)

water waves (m)

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waves in a wheat field (m)
3 0
2 years ago
Read 2 more answers
A square conducting loop 8.4 cm on a side is placed in a uniform B-field so that the plane of the loop is perpendicular to the d
arsen [322]

Answer:

Explanation:

area of square loop A = side²

= 8.4² x 10⁻⁴

A = 70.56 x 10⁻⁴ m²

when it is converted into rectangle , length = 14.7  , width = 2.1

area = length x width

= 14.7 x 2.1 x 10⁻⁴

= 30.87 x 10⁻⁴ m²

Let magnetic field be B

Change in flux = magnetic field x change in area

= B x ( 70.56 x 10⁻⁴ - 30.87 x 10⁻⁴ )

= 39.69 x 10⁻⁴ B

rate of change of flux = change in flux / time taken

= 39.69 x 10⁻⁴ B  / 6.5 x 10⁻³

= 6.1 x 10⁻¹ B

emf induced = 6.1 x 10⁻¹ B

6.1 x 10⁻¹ B  = 14.7 ( given )

B = 2.41 x 10

= 24.1 T

B ) magnetic flux is decreasing , so it needs to be increased as per Lenz's law . Hence current induced will be anticlockwise so that additional  magnetic flux is induced out of the page.

4 0
2 years ago
A solid spherical insulator has radius r = 2.5 cm, and carries a total positive charge q = 8 × 10-10 c distributed uniformly thr
ira [324]
At r = 2R> R The expression for the electric field will be given by: (2R)^2*E=kQ. Where, k=(9*10^9)N.m/C^2, Q=(8*10^-10)C and R=0.025m.  So substituting and clearing, we have that the magnitude of the electric field will be:  E=(9*10^9)*(8*10^-10)/((2*0.025)^2)=2880 N / C.
5 0
2 years ago
A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the
Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, (\theta) = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ F_x=Fcos(\theta)          and             ⇒ F_y=Fsin (\theta)

⇒ F_x=15cos(15)                           ⇒ F_y=15sin (15)

⇒ Applying concept of  forces.

⇒ \sum F_x=F_n_e_t =F-f

⇒ F_n_e_t =F-f

⇒ ma =F-f       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

⇒ a =\frac{15cos(15)-0}{40}     <em>...f is the friction which is zero here.</em>

⇒ a =\frac{14.48}{40}

⇒ a=0.362\ ms^-^2

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0
2 years ago
What is the wavelength of light waves if their frequency is 5.0x1014 Hz and the speed of light 3 x
Marat540 [252]

Answer:

D. 0.60 um

Explanation:

Given the following data;

Frequency = 5.0x10^14 Hz

Speed = 3 x 10^8 m/s

To find the wavelength;

Wavelength = speed/frequency

Substituting into the equation, we have

Wavelength = 3 x 10^8/5.0x10^14

Wavelength = 6 × 10^7m

But 1 meter (m) = 1000000 micrometer (um)

6 × 10^7 meter = 0.6 micrometer.

5 0
2 years ago
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