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kirza4 [7]
3 years ago
11

Quiz Review Problems

Physics
1 answer:
zlopas [31]3 years ago
5 0

The momentum of a neutron p = 586.25 kg m / s.

<u>Explanation:</u>

The product of mass and the velocity gives the momentum of an object and it is a vector quantity. It is denoted by the letter p. The unit of momentum is kilogram meter per second (or) kg m / s.

Given mass m = 1.675 \times 10,            velocity v = 3.500 \times 10

                  Momentum, p = mv

where m represents the mass,

          v represents the velocity.

                   momentum p = (1.675 \times 10) \times (3.500 \times 10)

                   momentum p = 586.25 kg m / s.

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What happens to the width of the central diffraction pattern (in the single slit experiment) as the slit width is changed and wh
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Answer:

width of fringes are increased

Explanation:

The width of central maxima is given by the following expression

Width = 2 x Dλ / d

  1. D is distance of screen from source , d is slit width and λ is wavelength of light source.
  2. Here we see , on d getting decreased , width will increase because d is in denominator .

Due to increased width ,  position of a fringe  moves away from the centre.

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What happens is a series circuit when you increase the number of bulbs?
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The bulbs will produce lesser light than their capacity, In short they will be dimmer because the the energy will get divided in the number of bulbs.

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Why is the surface salinity of the ocean higher in the subtropics than in equatorial regions?
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3 years ago
What is the nature of force between two charges if q1+q2=0
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Answer:
The force will be attractive.


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3 0
4 years ago
The position coordinate of a particle which is confined to move along a straight line is given by s =2t3−24t+6, where s is measu
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Answer:

a) the answer is t=4 seconds

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c) displacement= 142 m

Explanation:

Given the position of the particle

s=2t^3-24t+6

a) the time required when velocity v=72 m/s

v=72=\frac{ds}{st}=6t^2-24

now we solve for time t

6t^2=72+24=96\\\\\implies t^2=\frac{96}{6}\\\therefore t=4 s

b) acceleration when v=30 m/s

acceleration is the time derivative of velocity i.e

a=\frac{dv}{dt}=\frac{d}{dt}(30)=0

c) the net displacement of the particle during the interval t = 1 s to t = 4 s is

s_4-s_1=2t^3-24+6|_{t=4}-2t^3-24+6|_{t=1}\\s_4-s_1=126-(-16)=142 m

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4 years ago
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