Answer 1) The electric field at distance r from the thread is radial and has magnitude
E = λ / (2 π ε° r)
The electric field from the point charge usually is observed to follow coulomb's law:
E = Q / (4 π ε° 
)
Now, adding the two field vectors:
= {2.5 / (22 π ε° X 0.07 ) ; 0}
Answer 2) 
= {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))
Adding these two vectors will give the length which is magnitude of the combined field.
The y-component / x-component gives the tangent of the angle with the positive x-axes.
Please refer the graph and the attachment for better understanding.
Newton’s Law: F = MA
A = F/M (change equation)
12.6 N/ 2.4 kg = 5.25
Answer: acceleration is 5.25 m/s^2
One of the many random useless factoids that I carry around
in my head is the factoid that 60 miles per hour is equivalent
to exactly 88 feet per second.
So in three seconds at that speed, you would cover exactly
(3 x 88) = 264 feet.
Answer:
The the analysis for the free fall part should be done under the constant acceleration.
Explanation:
In the given problem, the jumper is falling under the free fall. Since, no external force is acting on the body therefore, the fall will be under the action gravity only. also, the acceleration due to gravity is always constant.
Therefore, the the analysis for the free fall part should be done under the constant acceleration.
Answer:
Explanation:
<u>Accelerated Motion
</u>
When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by
where a is the acceleration, and vo is the initial speed
.
The train has two different types of motion. It first starts from rest and has a constant acceleration of 
for 182 seconds. Then it brakes with a constant acceleration of 
until it comes to a stop. We need to find the total distance traveled.
The equation for the distance is
Our data is
Let's compute the first distance X1
Now, we find the speed at the end of the first period of time
That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0
Computing the second distance
The total distance is
