There are two particular cases, the first is when Object A is attracted to the neutral wall. This would indicate that the object is not neutral, as there is an attraction.
At the same time we know that Object A is attracted to an object B. And therefore, the load of A must be opposite to that of B. Remember that opposite charges attract each other. If the charge of object B is positive, then the charge of object A will be negative.
Option B is correct: It has a negative charge.
Answer:
the color .............................................................................................
Explanation:
Answer:
−8.59 m/s.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = −4.0 m/s
Acceleration (a) = −0.27 m/s²
Time (t) = 17 s
Final velocity (v) =?
Acceleration is defined as the change of velocity with time. Mathematically, it is expressed as:
Acceleration = change of velocity / time
Acceleration = (final velocity – Initial velocity) / time
a = (v – u) / t
With the above formula, we can determine the velocity of the train after 17 s. This can be obtained as follow:
Initial velocity (u) = −4.0 m/s
Acceleration (a) = −0.27 m/s²
Time (t) = 17 s
Final velocity (v) =?
a = (v – u) / t
−0.27 = (v – –4 ) / 17
−0.27 = (v + 4 ) / 17
Cross multiply
−0.27 × 17 = v + 4
−4.59 = v + 4
Collect like terms
−4.59 – 4 = v
−8.59 = v
v = −8.59 m/s
Thus, the velocity of the train after 17 s is −8.59 m/s.
What it looks to be that you found in A was the "initial"...b/c the question asks:
<span>"how much energy does the electron have 'initially' in the n=4 excited state?" </span>
<span>"final" would be where it 'finally' ends up at, ie. its last stop...as for this question...the 'ground state' as in its lowest energy level. </span>
The answer comes to: <span>−1.36×10^−19 J</span>
You use the same equation for the second part as for part a.
<span>just have to subract the 2 as in the only diff for part 2 is that you use 1squared rather than 4squared & subract "final -initial" & you should get -2.05*10^-18 as your answer. </span>
Answer:
(a) when the distance x is 0.3 m, the boundary layer thickness is 0.0055 m = 5.5 mm.
(b) when the distance x is 3 m, the boundary layer thickness is 0.0174 m = 17.4 mm.
(c) when the distance x is 30 m, the boundary layer thickness is 0.055 m = 55 mm.
Explanation:
For a laminar flow:
;
where;
d is the boundary layer thickness = 11 mm
x is the distance from the leading edge = 1.2 m
C is a constant =?
<u>Part (a):</u> when x = 0.3m
d = C√x
= 0.010042 × √0.3
= 0.0055 m = 5.5 mm
<u>Part (b):</u> when x = 3.0 m
d = C√x
= 0.010042 × √3.0
= 0.0174 m = 17.4 mm
<u>Part (c):</u> when x = 30 m
d = C√x
= 0.010042 × √30
= 0.055 m = 55 mm