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Andru [333]
3 years ago
13

An electrical power plant manages to send 88% of the heat produced in the burning of fossil fuel into the water-to-steam convers

ion. Of the heat carried by the steam, 40% is converted to the mechanical energy of the spinning turbine. Which of the following choices best describes thvall efficiency of the heat-to-work conversion in the plant (as a percentage)?
A. less than 40%
B. exactly 147%
C. 64%
D. greater than 88%
E. 40%
Physics
1 answer:
Naya [18.7K]3 years ago
4 0

Answer:

A

Explanation:

Let the x represent the amount of heat generated from the fossil fuel.

88% of x = 0.88 x

0.88 x was used to convert water to steam.

heat carried by steam = 40% × 0.88 x = 0.352 x

efficiency of the heat -to- work conversion = work output / work input = 0.352 x / x = 0.352 × 100 = 35.2 % which is less than 40 %

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A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3° angle. What i
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v_{1fy} = - 0.4549 m / s

Explanation:

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3 years ago
A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th
photoshop1234 [79]

Answer:

The height is  h_c = 42.857

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is h_s = 30 \ cm

            The angle of the slope is \theta = 15^o

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

                          mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2

Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    I = \frac{2}{5} mr^2

  The angular velocity w is mathematically represented as

                         w = \frac{v}{r}

So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

             mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]

            gh_s =[\frac{7}{10} ] v^2

              v^2 = \frac{10gh_s}{7}

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

             I = mr^2

Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 mgh_c  = mv^2

                     gh_c = v^2

                     h_c = \frac{v^2}{g}

Recall that   v^2 = \frac{10gh_s}{7}

                    h_c= \frac{\frac{10gh_s}{7} }{g}

                      = \frac{10h_s}{7}

            Substituting values

                   h_c = \frac{10(30)}{7}

                       h_c = 42.86 \ cm    

       

     

                         

8 0
2 years ago
A car accelerates from rest at a constant rate of 2m/s2 for 5s. what is the speed of a car at the end of that time?
Fiesta28 [93]
Good afternoon.


We have:

\mathsf{V_0 = 0}\\ \mathsf{a = 2 \ m/s^2}\\ \mathsf{t = 5 \ s}

The function of velocity:

\mathsf{V = V_0+at}\\ \\ \mathsf{V = 0 + 2t}\\ \\ \mathsf{V = 2t}


For t = 5 s:

\mathsf{V = 2\cdot 5}\\ \\ \boxed{\mathsf{V = 10 \ m/s}}
4 0
3 years ago
A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie
Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, V_{Hall} = 20\ mV = 0.02\ V

Magnetic Field, B = 0.10 T

Hall emf, V'_{Hall} = 69\ mV = 0.069\ V

Now,

Drift velocity, v_{d} = \frac{V_{Hall}}{B}

v_{d} = \frac{0.02}{0.10} = 0.2\ m/s

Now, the expression for the electric field is given by:

E_{Hall} = Bv_{d}sin\theta                            (1)

And

E_{Hall} = V_{Hall}d

Thus eqn (1) becomes

V_{Hall}d = dBv_{d}sin\theta

where

d = distance

B = \frac{V_{Hall}}{v_{d}sin\theta}                      (2)

(a) When \theta = 90^{\circ}

B = \frac{0.069}{0.2\times sin90} = 0.345\ T

(b) When \theta = 60^{\circ}

B = \frac{0.069}{0.2\times sin60} = 0.398\ T

5 0
2 years ago
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