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Nata [24]
3 years ago
14

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

the surroundings). Given these two temperatures, what is the maximum possible efficiency the car can have? (Note that actual car engine efficiencies are in the 20-25% range.) _______ %.
Physics
1 answer:
Step2247 [10]3 years ago
4 0

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%

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CaHeK987 [17]
It is A or D but I believe A
5 0
3 years ago
The diameter of a 12-gauge copper wire is 0.081 in. The maximum safe current it can 17) carry (in order to prevent fire danger i
Soloha48 [4]

Answer:

0.44m/s

Explanation:

drift velocity=I/nAq

diameter 12 gauge

wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*

V = 0.44m/s

6 0
4 years ago
Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulle
tangare [24]

Answer:

F=1.159

Explanation:

From the question we are told that:

Mass of pulley M=1kg

Radius r=12cm

Mass of block A M_a=2.1kg

Mass of block B m_b=4.1kg

Spring constant\mu= 358 J/m2

Generally the equation for Torque is mathematically given by

Since \sumF=ma

At mass A

 T_2-f_3=2.1a

At mass B

 4.8-T_1=4.1a

At  Pulley

 R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}

 R(T_1-T_2)=0.55a

Therefore the equation for total force F

At mass A+At mass B+At  Pulley

 (T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a

 (T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a

 -f_3+4.1=6.75a

 -f_3=6.75a+4.8

Since From above equation

M_{eff}=6.7kg

Therefore

T=2\pi \sqrt{{\frac{M_{eff}}{k}}

T=2\pi \sqrt{{\frac{6.75}{\mu}}

T=0.862s

Generally the equation for frequency is mathematically given by

F=\frac{1}{T} \\F=\frac{1}{0.862}

F=1.159

3 0
3 years ago
An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2
Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

M=-\frac{q}{p}

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

M=-\frac{200 cm}{50 cm}=-4.0

7 0
3 years ago
What is an expression for the vertical component of this vector?
grandymaker [24]
The correct choice is D .
4 0
3 years ago
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