An automobile fuel tank is filled to the brim with 45 L of gasoline (12 gal) at 10°C. Immediately afterward, the vehicle is park
ed in the sunlight, where the temperature is 35°C. How much gasoline overflows from the tank as a result of the expansion? (Neglect the expansion of the tank.)
1 answer:
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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
Answer:
-1.43 m/s relative to the shore
Explanation:
Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:
where are the mass of the swimmer and raft, respectively.
are the velocities of the swimmer and the raft after the run, respectively. We can solve for
So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore