A droplet of ink in an industrial ink-jet printer carries a charge of 1.6 x 10–11 coulombs and is deflected onto paper by a forc
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Answer:
R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks
Explanation:
We apply Newton's second law:
∑F=m*a
velocity is constant ,then , a=0
Nomenclature
W: weight
m: mass
N : normal force
Ff: Friction force
μk: coefficient of kinetic friction
T: tension force in the rope
F: applied horizontal force
g: acceleration due to gravity.
Force Calculation
W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N
W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N
∑Fy=0
N₁-W₁=0 , N₁=W₁ = 52.92 N
N₂-W₂=0, N₂=W₂=80.36N
Ff₁= μk* N₁=0.4*52.92 N = 21.16N
Ff₂= μk* N₂=0.4*80.36N = 32.14N
Look at the attached graphic
Free-Body diagram m₁=5.4 kg
∑Fx=0
T- Ff₁=0 , T= Ff₁ , T= 21.16N
Free-Body diagram m₂=8.2 kg
∑Fx=0
F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N
Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)
R= F/T= 53.3N/21.16N = 2.5
The elements found in the Group 1, or the Alkali Metal Group, have electronic configurations that end in 
.This means that they have 1 electron readily available to release in order to achieve a stable state.
When these atoms release the valence electron, they will achieve a stable state. For example, Lithium's stable state will be
and Sodium will be 
.
The oxidation state will then be +1.
The answer is C.
When these atoms release the valence electron, they will achieve a stable state. For example, Lithium's stable state will be
The oxidation state will then be +1.
The answer is C.
The acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.
Answer:
Explanation:
As per Newton's second law of motion, acceleration of any object is directly proportional to the net external unbalanced force acting on that object and inversely proportional to the mass of the object.
Since there are two forces acting on the box in opposite direction, the net force will be the difference of horizontal and frictional force acting on the object.
Net force = Horizontal force - Frictional force = 18 N - 8 N = 10 N.
Now, from second law of motion,
So, acceleration = 10 N /5 kg = 2 m/s².
Since, acceleration exerted by the box is found to be 2 m/s², we can determine the final velocity of the object after 10 seconds using the first equation of motion.
v = u + at, Here v is the final velocity and u is the initial velocity which is zero for the present case. Other parameters like a is found to be 2 m/s² and time is 10 seconds.
So Final velocity v = 0+(2×10)=20 m/s.
And the final position can be determined using the second equation of motion.
s = ut+1/2at²
Final position = (0×10)+(0.5×2×10×10)= 100 m.
So the final position is 100 m.
Thus, the acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.