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3 years ago

If a car's engine gives off 65% thermal energy, what is the maximum efficiency?

Physics

1 answer:

stich3 [128]3 years ago

5 0

Answer:

35%

Explanation:

The car's engine gives off 65% thermal energy

So only 35 % is converted into mechanical energy .

input heat = Q₁ = 100

output heat = Q₂ = 65

Work output = Q₁ - Q₂ = W

W = 100 - 65 = 35

Efficiency = W / Q₁ X 100

= (35/ 100) X 100

= 35%.

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a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T

wlad13 [49]

Explanation:

Given: $F=m\ddot{x}=Fe^{-\frac{t}{T}}$

Solving for $\ddot{x}$ :

$\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}$

where:

$T=\sqrt{\frac{m}{F}}$

Integrating to get $\dot{x}$ with initial conditions $\dot{x}(0)=0$ :

$\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}$

Integrating to get x with initial conditions x(0) = 0:

$x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}$

When t=T:

$x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}$

$\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\sqrt{\frac{F}{m}}(1-\frac{1}{e})$

4 0

3 years ago

If Star A and Star B have the same absolute magnitude, but Star A is brighter, what does that tell us?

Hatshy [7]

We can conclude that star A is closer to us than star B.

In fact, the absolute magnitude gives a measure of the brightness of the star, if all the stars are placed at the same distance from Earth. So, it's a measure of the absolute luminosity of the star, indipendently from its distance from us: since the two stars have same absolute magnitude, it means that if they were at same distance from Earth, they would appear with same luminosity. Instead, we see star A brighter than star B, and the only explanation is that star A is closer to Earth than star B (the closer the star A, the brigther it is)

6 0

4 years ago

A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0

ankoles [38]

Answer:

a. B = 0.20T

b. u = 17230.6 J/m³

c. E = 0.236J

d. L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

$B=\frac{\mu_o n i}{L}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current = 90.0A

You replace the values of the parameters in the equation (1):

$B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T$

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

$u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}$

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

$E=uV$ (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

$V=AL$

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

$V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3$

Then, the energy contained in the solenoid is:

$E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J$

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

$L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H$

The inductance of the solenoid is 5.84*10^-5 H

3 0

4 years ago

What is the area of the bottom of a tank 30.0 cm long and 15.0 cm wide?

Igoryamba

Answer:

perimeter of the bottom of the tank is 450 cm

Explanation:

If you want to get area you need the Length, width,and height.

5 0

3 years ago

Read 2 more answers

A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of

Step2247 [10]

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

$\eta = 1-\frac{T_L}{T_H}$

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

$\eta = 1-\frac{T_L}{T_H}$

$\eta = 1-\frac{293}{713}$

$\eta = 0.589$

Therefore the maximum possible efficiency the car can have is 58.9%

4 0

3 years ago