Diesel engines give more miles per gallon than gasoline engines, but some of this is due to the higher energy content of diesel
1 answer:
Answer:
0.106
Explanation:
For 1 liter of diesel the car can get 19 km, if it takes 0.2 MJ for each km then it would take the total energy of 19*0.2 = 3.8 MJ to move an aerodynamic car 19 km. Since 1 liter of of diesel also contains 36 MJ in internal energy, then the efficiency of the diesel engine is the ratio of its output energy over its input energy:
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1) At x = 6.6 cm,
2) At x = 6.6 cm,
3) At x = 1.45 cm,
4) At x = 1.45 cm,
5) Surface charge density at b = 4 cm:
6) At x = 3.34 cm, the x-component of the electric field is zero
7) Surface charge density at a = 2.9 cm:
8) None of these regions
Explanation:
1)
The electric field of an infinite sheet of charge is perpendicular to the sheet:
where
is the surface charge density
is the vacuum permittivity
The field produced by a thick slab, outside the slab itself, is the same as an infinite sheet.
So, the electric field at x = 6.6 cm (which is on the right of both the sheet and the slab) is the superposition of the fields produced by the sheet and by the slab:
where
The field of the sheet is to the left (negative charge, inward field), while the field of the slab is the right (positive charge, outward field).
So,
And the negative sign indicates that the direction is to the right.
2)
We note that the field produced both by the sheet and by the slab is perpendicular to the sheet and the slab: so it is directed along the x-direction (no component along the y-direction).
So the total field along the y-direction is zero.
This is a consequence of the fact that both the sheet and the slab are infinite along the y-axis. This means that if we take a random point along the x-axis, the y-component of the field generated by an element of surface dS of the sheet (or the slab), , is equal and opposite to the y-component of the field generated by an element of surface dS of the sheet located at exactly on the opposite side with respect to the x-axis,
. Therefore, the net field along the y-direction is always zero.
3)
Here it is similar to part 1), but this time the point is located at
x = 1.45 cm
so between the sheet and the slab. This means that both the fields of the sheet and of the slab are to the left, because the slab is negatively charged (so the field is outward). Therefore, the total field is
Substituting the same expressions of part 1), we find
where the negative sign indicates that the direction is to the left.
4)
This part is similar to part 2). Since the field is always perpendicular to the slab and the sheet, it has no component along the y-axis, therefore the y-component of the electric field is zero.
5)
Here we note that the slab is conductive: this means that the charges in the slab are free to move.
We note that the net charge on the slab is positive: this means that there is an excess of positive charge overall. Also, since the sheet (on the left of the slab) is negatively charged, the positive charges migrate to the left end of the slab (at a = 2.9 cm) while the negative charges migrate to the right end (at b = 4 cm).
The net charge per unit area of the slab is
And this the average of the surface charge density on both sides of the slab, a and b:
(1)
Also, the infinite sheet located at x = 0, which has a negative charge , induces an opposite net charge on the left surface of the slab, so
(2)
Now we have two equations (1) and (2), so we can solve to find the surface charge densities on a and b, and we find:
6)
Here we want to calculate the value of the x-component of the electric field at
x = 3.34 cm
We notice that this point is located inside the slab, because its edges are at
a = 2.9 cm
b = 4.0 cm
But slab is conducting , and the electric field inside a conductor is always zero (because the charges are in equilibrium): therefore, this means that the x-component of the electric field inside the slab is zero
7)
We calculated the value of the charge per unit area on the surface of the slab at x = a = 2.9 cm in part 5), and it is
8)
As we said in part 6), the electric field inside a conductor is always zero. Since the slab in this problem is conducting, this means that the electric field inside the slab is zero: therefore, the regions where the field is zero is
2.9 cm < x < 4 cm
So the correct answer is
"none of these region"
Learn more about electric fields:
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Answer:
(a) Height is 4.47 m
(b) Height is 4.37 m
Solution:
As per the question:
Initial velocity of teh ball,
Angle made by the ramp,
Distance traveled by the ball on the ramp, d = 5.00 m
Now,
(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:
where
H =
g =
= 19.06 m/s
Now, maximum height attained is given by:
Height from the ground =
(b) now, considering the coefficient of friction bhetween ramp and the ball, :
velocity can be given by the eqn-3 of motion:
= 18.7 m/s
Now, maximum height attained is given by:
Height from the ground =
Answer:
Please find attached
Explanation:
Answer:
Explanation:
Here is the full question and answer,
The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.
When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.
Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.
Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?
Answer
A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.
Calculate the speed of the archer fish.
The time of the flight of spitted water is,
Substitute for g and
for
in above equation.
Spitted water will travel horizontally.
Displacement of water in this time period is
Substitute for
and
for x in above equation.
B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.
Calculate the height of insect above the surface.
Vertical component of the velocity is,
Substitute for v and
for
in above equation.
Horizontal component of the velocity is,
Substitute for v and
for
in above equation.
When horizontal distance away from the fish.
The time of flight for distance (d) is ,
Substitute for d and
for
in equation
Distance of the insect above the surface is,
Substitute for
and
for t and
for g in above equation.