Answer:
Final velocity of electron, 
Explanation:
It is given that,
Electric field, E = 1.55 N/C
Initial velocity at point A, 
We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :
........(1)
a is the acceleration, 
We know that electric force, F = qE
Use above equation in equation (1) as:
v = 647302.09 m/s
or
So, the final velocity of the electron when it reaches point B is 
. Hence, this is the required solution.
Answer:
Explanation:
KE = ½mv² = ½(6.8)8² = 217.6 J
round as appropriate because that result is way too much precision for the inputs provided. Arguably should be 200 J based on the single significant digit of the velocity.
A radio frequency identification application would most likely interface with an Operational Data Store.
The Operations Data Store (ODS) is a central database that provides the latest data snapshots from multiple transaction systems for operational reporting.
It allows organizations to combine data in its original format from various sources into a single destination to provide business reporting.
ODS contains integrated updates from operational sources and supports business intelligence (BI) tools to facilitate tactical decision making.
For example, an administrator can configure ODS to pull weekly batches of data from a billing application that is rarely updated, importing individual transaction records as they occur in the sales database(thanks to these database triggers), then combine the two into new relational tables.
As a result, querying and reporting on operational data in ODS comes with the assurance that these integrated tables contain the latest and most relevant snapshots of the business.
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(a) The acceleration of the system is determined as 1.58 m/s².
(b) The relative weight of P is pounds is determined as 0.14 lb.
<h3>
Acceleration of the system</h3>
The acceleration of the system is calculated as follows;
W - T = m₂a --- (1)
T = m₁a ----(2)
μmgsinθ - m₁a = m₂a
(0.3 x 3 x 9.8 x sin40) - (0.4 + 0.2)a = 3a
5.67 - 0.6a = 3a
5.67 = 3.6a
a = 5.67/3.6
a = 1.58 m/s²
<h3>
Relative Weight of P</h3>
W = ma
W = 0.4 x 1.58
W = 0.632 N = 0.14 lb
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<span>Fusion produces large amounts of energy, and the fuel is found on Earth.</span>
