Answer:
Explanation:
m = Mass of each the cars =
= Initial velocity of first car = 3.46 m/s
= Initial velocity of the other two cars = 1.4 m/s
v = Velocity of combined mass
As the momentum is conserved in the system we have
Speed of the three coupled cars after the collision is .
As energy in the system is conserved we have
The kinetic energy lost during the collision is .
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is
Explanation:
From the question we are told that
The diameter of the wire is
The radius of the wire is
The resistivity of aluminum is
The electric field change is mathematically defied as
Generally the charge is mathematically represented as
Where A is the area which is mathematically represented as
So
Therefore
substituting values
From the question we are told that t = 5 sec
Answer:
The combined velocity is 8.61 m/s.
Explanation:
Given that,
The mass of a truck, m = 2800 kg
Initial speed of truck, u = 12 m/s
The mass of a car, m' = 1100 kg
Initial speed of the car, u' = 0
We need to find the combined velocity the moment they stick together. Let it is V. Using the conservation of momentum.
So, the combined velocity is 8.61 m/s.
Answer:
a. cosθ b. E.A
Explanation:
a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for 90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function
b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A
