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krok68 [10]
3 years ago
7

A certain gasoline engine has an efficiency of 35.3%. What would the hot reservoir temperature be for a Carnot engine having tha

t efficiency, if it operates with a cold reservoir temperature of 160°C? (°C)
Physics
1 answer:
levacccp [35]3 years ago
4 0

Answer:

The temperature of hot reservoir is 669.24 K.

Explanation:

It is given that,

Efficiency of gasoline engine, \eta=35.3\%=0.353

Temperature of cold reservoir, T_C=160^{\circ}C=433\ K

We need to find the temperature of hot reservoir. The efficiency of Carnot engine is given by :

\eta=1-\dfrac{T_C}{T_H}

T_H=\dfrac{T_C}{1-\eta}

T_H=\dfrac{433}{1-0.353}

T_H=669.24\ K

So, the temperature of hot reservoir is 669.24 K. Hence, this is the required solution.

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It took 3.5 hours for a train to travel the distance between two cities at a velocity
Amiraneli [1.4K]

Answer:

420

Explanation:

420 because 120 x 3 = 360

120/2 = 60 and 360 + 60 = 420

I hope this helps you and please consider giving me brainliest :)

7 0
3 years ago
Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance
siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

       

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = k \frac{Q}{x_1^2} + k  \frac{Q}{x_2^2}

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = k  \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )

            E_ {total} = k \frac{Q}{x_1^2}   ( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

           

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = -k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = + k \frac{Q}{(d-x_1)^2}+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

             

5) Right side point

            E_ {total} = k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}

             E_ {total} = - k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0
2 years ago
Erica throws a tennis ball against a wall, and it bounces back. Which force is responsible for sending the ball back to Erica?
Anna71 [15]

OPTIONS :

A.) the force that the ball exerts on the wall

B.) the frictional force between the wall and the ball

C.) the acceleration of the ball as it approaches the wall

D.) the normal force that the wall exerts on the ball

Answer: D.) the normal force that the wall exerts on the ball

Explanation: The normal force acting on an object can be explained as a force experienced by an object when it comes in contact with a flat surface. The normal force acts perpendicular to the surface of contact.

In the scenario described above, Erica's tennis ball experiences an opposite reaction after hitting the wall.This is in relation to Newton's 3rd law of motion, which states that, For every action, there is an equal and opposite reaction.

The reaction force in this case is the normal force exerted on the ball by the wall perpendicular to the surface of contact.

8 0
3 years ago
How are cactus adapted to survive in deserts?
andreev551 [17]

1. They have evolved their leaves into spikes for minimum water loss through transpiration.

2. They have a waxy layer for minimum water loss.

3. They have thick walls for minimum water loss.

4. They can take water from atmosphere.

5. They change the photo energy from Sun into an intermediate stage and store it, so that they can make food even in night.

7 0
2 years ago
Suppose that a particular artillery piece has a range R = 9880 yards . Find its range in miles. Use the facts that 1mile=5280ft
Solnce55 [7]

Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

6 0
3 years ago
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