A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied o
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Answer:
Part a)
Velocity = 6.9 m/s
Part b)
Position = (3.6 m, 5.175 m)
Explanation:
Initial position of the particle is ORIGIN
also it initial speed is along +X direction given as
now the acceleration is given as
when particle reaches to its maximum x coordinate then its velocity in x direction will become zero
so we will have
Part a)
the velocity of the particle at this moment in Y direction is given as
Part b)
X coordinate of the particle at this time
Y coordinate of the particle at this time
so position is given as (3.6 m, 5.175 m)
Answer:
The spacing is 5.15 μm.
Explanation:
Given that,
Electron with energy = 25 eV
Wave length = 0.25 nm
Separation d= 0.16 mm
Distance D=3.3 m
We need to calculate the spacing
Using formula of width
Put the value into the formula
Hence, The spacing is 5.15 μm.