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lozanna [386]
3 years ago
10

An automobile engine delivers 47.4 hp. how much time will it take for the engine to do 6.82 × 105 j of work? one horsepower is e

qual to 746 watts. answer in units of s.
Physics
1 answer:
kogti [31]3 years ago
8 0
1 horsepower is equal to 746 W, so the power of the engine is
P=47.4 hp \cdot 746  \frac{W}{hp}=35360 W
The power is also defined as the energy E per unit of time t:
P= \frac{E}{t}
Where the energy corresponds to the work done by the engine, which is E=6.82 \cdot 10^5 J. Re-arranging the formula, we can calculate the time t needed to do this amount of work:
t =  \frac{E}{P}= \frac{6.82 \cdot 10^5 J}{35360 W}=19.3 s
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Answer:

Just as images are reflected from the surface of a mirror, light reflected from a smooth water surface also produced a clear image. ... Consequently, the outgoing rays are reflected at many different angles and the image is disrupted. Reflection from such a rough surface is called diffuse reflection and appears matte.

Explanation:

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During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in
andreyandreev [35.5K]

Answer:

2872.8 N

Explanation:

We have the following information

m =n72kg

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t = 0.95s.

From here we use the equation

Δy=1/2at2 in order to solve for the acceleration.

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=( 2x 18m)/(0.95s²)

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3 years ago
Student A states that when she sits down on a chair, she is exerting a force on the chair and that is all that happens. Student
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Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b
zmey [24]

Answer:

The maximum amount of work is  W = 1563.289 \ J

Explanation:

From  the question we are told that

   The temperature of the environment is  T = 280\ K

    The volume of container A is  V_A = 2 m^3

    Initially the number of moles  is  n = 1.2 \ moles

     The volume of container B is V_B = 3.5 \ m^3

     

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

             W =  P_A V_A  ln[ \frac{V_B}{V_A} ]

Now from the Ideal gas law

          P_A V_A =  nRT

So substituting for P_A V_A in the equation above

          W =  nRT ln [\frac{V_B}{V_A} ]

Where R is the gas constant with a values of  R =  8.314 \  J/mol

Substituting values we have that

            W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]

          W = 1563.289 \ J

8 0
3 years ago
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