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madreJ [45]
3 years ago
11

Any kind of wave spreads out after passing through a small enough gap in a barrier. This phenomenon is known as _________. a) di

ffraction b) double-slit c) interference d) refraction e) antireflection
Physics
2 answers:
fenix001 [56]3 years ago
7 0

the answer is a diffraction

bulgar [2K]3 years ago
6 0

Answer:

a) diffraction

Explanation:

Diffraction occurs when waves pass through small openings, around obstacles or sharp edges.When an opaque object is between the point of light and a screen, the border between the shaded and illuminated regions on the screen is not defined. A careful inspection of the scrubber shows that a small amount of light is diverted to the shaded region. The region outside the shadow contains bright and dark altered bands, where the intensity of the first band is brighter than the region of uniform illumination.

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213.49 in standard form is
Rudiy27

Answer:

21349/ 100

Explanation:

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3 years ago
All of the following statements about the pyramid of biomass are correct EXCEPT: a. Biomass is the total dry mass of the organis
Goshia [24]

Answer: The answer is option B

Explanation:

The base of the pyramid represents primary producers since they help to pass on energy to consumers as we go higher in the pyramid. They are not consumers because they supply energy and do not use up energy. They are called autotrophs.

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Define heat according to molecular theory .​
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3 years ago
: The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is subjected to the torqu
Dominik [7]

Answer:

the maximum shear stress in the bar = 9.55 MPa

the twisting angle = 0.146^0

Explanation:

From the diagram below;

Torque (T) = 2 kN × 5 mm +  2 kN × 5 mm

= 2*10^3 *0.05 + 2*10^3 *0.05

= 200 N.m

Now; If we divide the shaft into two parts AC & BC

Then :

T = T_1 + T_2 = 200          ------------ equation (1)

At the junction :

\phi _{AC} = \phi _{BC}

\phi = \frac{Tl}{GJ}

⇒ \frac{T_1 \ l_{AC}}{GJ_1}=  \frac{T_2 \ l_{BC}}{GJ_2}

T_1l_{AC} = T_2 l_{BC}

T_1*400 = T_2 * 600\\T_1 = \frac {600T_2}{400}\\T_1 = 1.5 T_2

replace T_1 = 1.5 T_2 into above equation (1)

1.5 T_2 +T_2 = 200\\2.5 T_2 = 200\\T_2 = \frac{200}{2.5}\\T_2 = 80 N.m

Again:

T_1 +T_2 = 200\\T_1 + 80 = 200\\T_1 = 200 - 80\\T_1 = 120 N.m

Now ; we can deduce that the maximum shear comes from \phi_{AC} since T_1 = 120 \ N.m

So;

\gamma_{max} = \frac{T_1 *R}{ \frac {\pi D^4}{32}}

where;

R = 20 mm

D = 40 mm = 0.04 m

Then;

\gamma_{max} = \frac{120 *20*10^{-3}}{ \frac {\pi (0.04)^4}{32}}

\gamma_{max} = 9.55 \ MPa

Thus ; the maximum shear stress in the bar = 9.55 MPa

b)

Again: \phi_c = \frac{T_1l_{AC}}{JG} =\frac{T_2l_{AC}}{JG}

\phi_c = \frac{120*0.4}{75*10^9*\pi *\frac{(0.04)^2}{32}}

\phi_c = 2.55 *10^{-3} \ rads

\phi_c = 0.146^0

Thus, the twisting angle = 0.146^0

4 0
3 years ago
When an object is located 32 cm to the left of the lens, the image is formed 17 cm to the right of the lens. What is the focal l
Harman [31]

Answer: 11.1cm

Explanation:

Object distance (u) = 32cm

Image distance(v) = 17cm

Focal length(f) =?

Lens formular:

1/f = 1/u + 1/v

1/f = 1/32 + 1/17

Taking the L. C. M of 17 and 32

1/f = (17 + 32) / 544

1/f = 49/544

Taking the reciprocal

f = 544/49

f = 11.1 cm

Therefore, Focal length of the lens is 11.1cm

7 0
4 years ago
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