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lapo4ka [179]
2 years ago
15

The average velocity of an object over 6.0 seconds interval is 2 m/s what is the total distance traveled and M by the object doi

ng this time interval
Physics
1 answer:
lina2011 [118]2 years ago
8 0

Answer:

<h2>The answer is 12 m</h2>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

distance = velocity × time

From the question we have

distance = 2 × 6

We have the final answer as

<h3>12 m</h3>

Hope this helps you

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Explain how the aperture geometry relates to the<br> diffraction pattern.
konstantin123 [22]

Answer:

The answer to this question is given below in the explanation section.

Explanation:

how the aperture geometry relates to the

diffraction pattern:

Diffraction is the spreading out of waves as they pass through an aperture or around objects.it occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of  the incident wave.For every small aperture sizes,the vast majority of the wave is blocked.For large apertures the wave passes by or through the obstacle without any significant of diffraction.

            in an aperture with width smaller than the wavelength,the wave transmitted through the aperture spreads all the way around the behave like a point sources of waves.

                             single slit diffraction pattern

The diffraction pattern made by waves passing through a slit of width \alpha (larger than∫) can be understood by imagining a series of point sources all in phase along the width of the slit.The waves moving directly forward are all in phase,so they from a large central maximum.

                 if the waves travels at an angle Ф from the normal to the slit,then there is a path difference x between the waves production at the two end of the slit.

           x=a sinΦ

The path difference between the top and middle waves is λ/2 then they are exactly out of phase and cancel each other out. This happens to all consecutive pairs of waves (the ones produced by the second source from the top and the second source past the middle etc.)at the angle so there is no resultant wave at this angle.Thus a minimum is the diffraction pattern is obtained at

                                                   λ=α sinθ

Now slit can be divided into four equal sections and the pairing of sources to give destructive interference can be repeated for the top two section ,which is identical to the result of pairing off matching sources in the bottom two sections.in this case we obtained from the minimum.

             λ/2=α/4 sinθ

we can divided the slit aperture into six equal sections and pair off sources in the top two divisions and then the bottom two,to give destructive interference for every matched pair.The minimum of intensity are obtained at angles

                      nλ = α sinθ

where n is an integer (1,2,......),  but not n=0.There is a maximum of intensity in the center of the pattern. This process only gives the position of the minima,does not work for positions of the maxima,and so does not give the intensities of the maxima.

5 0
2 years ago
A block of mass m1=3.7kg on a smooth inclined plane of angle 30 is connected by a cord over a small frictionless pulley to a sec
kari74 [83]

consider the forces on mass m₁ on the incline plane :

parallel to incline , force equation is given as

T - m₁ g Sin30 = m₁ a

T = m₁ g Sin30 + m₁ a                                       eq-1

consider the force on mass m₂ on the incline plane :

m₂ g - T = m₂ a  

T = m₂ g - m₂ a                                                  eq-2

Using eq-1  and eq-2

m₂ g - m₂ a = m₁ g Sin30 + m₁ a

inserting the values

(2.3 x 9.8) - 2.3 a = (3.7 x 9.8) Sin30 + 3.7 a

a = 0.74 m/s²


3 0
3 years ago
4. What type of assessments are based on repeatable, measurable data?
Harman [31]

Answer:

  • OBJECTIVE ASSESSMENTS
  • SUBJECTIVE ASSESSMENTS
  • BODY COMPOSITION ASSESSMENTS
  • COGNITIVE ASSESSMENTS

Explanation:

PLS MAKE ME BRAINLIEST

8 0
2 years ago
22. State any three features of the electroscope.​
MatroZZZ [7]
-1- was created in the 1600 by william gilbert
-2-When the charge is positive, electrons in the metal of the electroscope are attracted to the charge and move upward out of the leaves. This results in the leaves to have a temporary positive charge and because like charges repel, the leaves separate. When the charge is removed, the electrons return to their original positions and the leaves relax
3-

An electroscope is made up of a metal detector knob on top which is connected to a pair of metal leaves hanging from the bottom of the connecting rod. When no charge is present the metals leaves hang loosely downward. But, when an object with a charge is brought near an electroscope, one of the two things can happen.
7 0
2 years ago
Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th
Mariulka [41]

Answer:

a) 0.0288 grams

b) 2.6*10^{-10} J/kg

Explanation:

Given that:

A typical human  body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance  for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

a)

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

b)

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = \frac{energy \ absorbed }{mass \ of \ the \ body}

= \frac{1.10*10^6*1.6*10^{-14}}{80}

= 2.2*10^{-10} \ J/kg

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) =  1.2  *2.2*10^{-10} \ J/kg

Effective dose (Sv) = 2.6*10^{-10} J/kg

 

5 0
3 years ago
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