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ehidna [41]
2 years ago
7

Diesel engines give more miles per gallon than gasoline engines, but some of this is due to the higher energy content of diesel

fuel. At highway speeds, it takes 0.20 MJ to move an aerodynamic car 1.0 km. At highway speeds, with a gasoline engine, a car gets 16 km per liter of fuel; with a diesel engine under the hood, the car gets 19 km per liter of fuel. One liter of diesel contains 36 MJ; 1 liter of gasoline contains only 32 MJ. Part A What is the efficiency of the diesel engine
Physics
1 answer:
Lina20 [59]2 years ago
5 0

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:

\frac{3.8}{36} = 0.106

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If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.55 x 10-4 T) at a distance of 25
antiseptic1488 [7]

we are given in the problem the following dimensions or specifications 
B = 0.000055 T r = 0.25 m constant mu0 = 4*pi*10-7 

The formula that is applicable from physics is 
B = mu0*I/(2*pi*r) I = 2*B*pi*r/mu0 I = 68.75 Amperes 
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The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. Which of the velocity components (n
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Neglecting air resistance, the horizontal component remains constant. The angle doesn't matter.
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Which statements describe how chemical formulas, such as H2O, represent compounds? Check all that apply.
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The statements that apply in this case are:

They show the elements that make up a compound.

They show the types of atoms that make up a molecule.

They show the number of each type of atom in a molecule.

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Three beads are placed on the vertices of an equilateral triangle of side d = 3.4cm. The first bead of mass m1=140gis placed on
Vlad [161]

Answer:

Xcm = 1.95 cm  and Ycm = 1.76 cm

Explanation:

The very useful concept of mass center is

     R cm = 1/M  ∑ m_{i}  r_{i}

Where ri, mi are the mass positions of the bodies from some reference point by selecting and M is the total mass of the body.

Let's look for the total mass

     M = m₁ + m₂ + m₃

     M = 140 + 45 + 85

     M = 270 g

Let's look for the position of each point

Point 1. top vertex, if the triangle has as side d

      R₁ = d / 2 i ^ + d j ^

      R₁ = (1.7 cm i ^ + 3.4 j ^) cm

Point 2. left vertex. What is the origin of the system?

      R₂ = 0

Point 3. Right vertex

      R₃ = d i ^

      R₃ = 3.4 i ^ cm

a) The x component of the massage center

      Xcm = 1 / M (m₁ x₁ + m₂ x₂ + m₃ x₃)

      Xcm = 1 / M (m₁ d / 2 + 0 + m₃ d)

      Xcm = d / M (m₁ / 2 + m₃)

b)   Let's write the mass center component x

      Xcm = 1/270 (1.7 140 + 0 + 3.4 85)

      Xcm = 238/270

      Xcm = 1.95 cm

c) let's find the component and center of mass

     Ycm = 1 / M (m₁ y₁ + m₂ y₂ + m₃ y₃)

    Ycm = 1 / M (m₁ d + 0 + 0)

    Ycm = m₁ / M d

d) let's calculate

    Y cm = 1/270 (140 3.4 + 0 + 0)

    Ycm = 1.76 cm

8 0
2 years ago
Suppose you throw a baseball downward from a roof so that it initially has 120 J of gravitational potential energy, and 10 J of
vlabodo [156]

Answer:

B.

It will be greater than 10 J.

Explanation:

The total mechanical energy of an object is the sum of its potential energy (PE) and its kinetic energy (KE):

E = PE + KE

According to the law of conservation of energy, when there are no frictional forces on an object, its mechanical energy is conserved.

The potential energy PE is the energy due to the position of the object: the highest the object above the ground, the highest its PE.

The kinetic energy KE is the energy due to the motion of the object: the highest its speed, the largest its KE.

Here at the beginning, when it is at the top of the roof, the baseball has:

PE = 120 J

KE = 10 J

So the total energy is

E = 120 + 10 = 130 J

As the ball falls down, its potential energy decreases, since its height decreases; as a result, since the total energy must remain constant, its kinetic energy increases (as its speed increases).

Therefore, when the ball reaches the ground, its kinetic energy must be greater than 10 J.

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