An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v
1 answer:
We are given: Final velocity ()=20 m/s .
Time t= 2.51 s and
distance s = 82.9 m.
We know, equation of motion
Let us plug values of final velocity, and time in above equation.
Subtracting 2.51a from both sides, we get
-----------equation(1)
Using another equation of motion
Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.
We get,
Now, we need to solve it for a.
20-20+2.51a=165.8a.
-163.29a=0
a=0.
So, the acceleration would be 0 m/s^2.
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Answer:
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Explanation:
Given :
De Broglie wavelength ,
Plank's constant ,
Charge of electron ,
Mass of electron , m=9.11\times 10^{-31}\ kg.
We know , according to de broglie equation :
Now , we know potential energy applied on electron will be equal to its kinetic energy .
Therefore ,
Putting all values in above equation we get ,
Hence , this is the required solution.
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 × m
dynamic viscosity = 1.75 × Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ
so
= µ ............1
put here value
= 1.75× ×
= 0.175 v
and
area between air and puck is given by
Area =
area =
area = 7.85 × m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 ×
force = 1.374 × v
and now apply newton second law
force = mass × acceleration
- force =
- 1.374 × v =
t =
time = 2.18
so time required after impact for a puck is 2.18 seconds