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KatRina [158]
3 years ago
10

a train moving west with an initial velocity of 20m/s accelerates 4m/s2 for 10 seconds . during this time , the train moves a di

stance of
Physics
1 answer:
Blababa [14]3 years ago
8 0
X=1/2at^2+vt+x0
x=1/2(4)(100)+20(10) = 400 m
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Which is the smallest atom? A. Magnesium (Mg) B. Calcium (Ca) C. Barium (Ba) D. Strontium (Sr)
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Try option B or option A

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A screen is placed 1.60 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40 c
Lunna [17]

Answer:

distance between the two second-order minima is 2.8 cm

Explanation:

Given data

distance = 1.60 m

central maximum = 1.40 cm

first-order diffraction minima = 1.40 cm

to find out

distance between the two second-order minima

solution

we know that fringe width = first-order diffraction minima /2

fringe width = 1.40 /2 = 0.7 cm

and

we know fringe width of first order we calculate slit d

β1 = m1λD/d

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and

fringe width of second order

β2 = m2λD/d

β2 = m2λD / ( m1λD/β1 )

β2 = ( m2 / m1 ) β1

we know the two first-order diffraction minima are separated by 1.40 cm

so

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put here value

y = 2 ( 2 / 1 ) 0.7

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so distance between the two second-order minima is 2.8 cm

6 0
3 years ago
The displacement (in meters) of a particle moving in a straight line is given by the equation of motion:
lutik1710 [3]

Answer:

  • At t = 1\; \rm s, the particle should have a velocity of -8\; \rm m \cdot s^{-1}.
  • At t = 2\; \rm s, the particle should have a velocity of -1\; \rm m \cdot s^{-1}.
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For a > 0, at t = a \; \text{second}, the particle should have a velocity of \displaystyle -\frac{8}{a^3}\; \rm m \cdot s^{-1}.

Explanation:

Differentiate the displacement of an object (with respect to time) to find the object's velocity.

Note that the in this question, the expression for displacement is undefined (and not differentiable) when t is equal to zero. For t > 0:

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This expression can then be evaluated at t = 1, t = 2, and t = 3 to obtain the required results.

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