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2 years ago

1) Calculate the ideal efficiency of a heat engine that takes in energy at 800 K and expels heat to a

Physics

1 answer:

masha68 [24]2 years ago

3 0

Answer:

Could you explain that more better?

Explanation:

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The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

$2M_2V_1^2=0.5\times M_2V_2^2$

$2V_1^2=0.5\times V_2^2$

$4V_1^2= V_2^2$

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

$2(V_1+9.0)^2=(2V_1+9.0)^2$

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

$(0.404V_1)=(3.72)$

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

8 0

2 years ago

Read 2 more answers

The Biot-Savart Law describes the relationship between which of the following? Select all that apply.

faltersainse [42]

The correct answer is magnetic field, electric field, and charges.

8 0

2 years ago

The scale is 1:30 and their are 10 boulders. How many boulders are their in the real pond

Alenkasestr [34]

Answer:

the correct answer is C

Explanation:

When we express that the scale is 1:30 we mean that the objects of the realization are reduced by a factor of 30 in the graph, for example a distance of 30 cm in the graph is represented by a distance of 1 cm.

Therefore something that in the graph has n value to bring it to real size must be multiplied by the scale.

Applying this to our case if there is

10 boulder on the chart

in reality there are #_boulder = 10 30

#_boulder = 300 boulder

so the correct answer is C

7 0

2 years ago

Why can’t a real machine ever have 100% efficiency

Harman [31]

Answer:

Almost all machines require energy to offset the effects of gravity, friction, and air/wind resistance. Thus, no machine can continually operate at 100 percent efficiency.

4 0

2 years ago

How do you do this?Or what are the formulus

LuckyWell [14K]

You got the formulas on the sheet on the top :) So just use those, exchanging v (as in velocity, expressed in m/s) and the d (in meters) and t (in seconds). Hope you will manage it.

4 0

2 years ago