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tensa zangetsu [6.8K]
3 years ago
9

What is the formula for calculating the efficiency of a heat engine

Physics
1 answer:
horsena [70]3 years ago
5 0
The answer is Eficiency=T<< Tox100
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An 85-kg refrigerator is located on the 70th floor of a skyscraper (4km above the ground ) what is the potential energy
Airida [17]
Potential energy can be calculated using the following rule:
potential energy = mgh where:
m is the mass = 85 kg
g is the acceleration due to gravity = 9.8 m/sec^2
h is the height = 4 km = 4000 meters

Substitute in the above equation to get the potential energy as follows:
Potential energy = 85*9.8*4000 = 3332000 joules
7 0
3 years ago
 5) You did 170 joules of work lifting a 140 N backpack. How high did you lift the backpack?          
xeze [42]
Sorry im from the UK, ill do my math in metric  :) 
5) work done = force * distance moved in direction of force

170J = 140N* ?d
170J/140N = 1.21m 

F=Ma

140N/9.81ms^-2=

14.27Kg or 31.46Ib

you could do the fast alternative simplified route to this question 

as 140/4.45 = 31.46Ib

5 0
3 years ago
Read 2 more answers
What is the electrical consumption in KVA of a motor powered by a 3-phase, 60 Hz, 460 VAC supply that continuously draws 17 A
MrRa [10]

Answer:

15.34 kVA

Explanation:

A motor is a device that converts electrical energy into mechanical energy. It takes in electrical energy at the input and produce torque (motion) at the output.

The power consumption for a three phase motor is the product of voltage and current and √3. The √3 is because it is a three phase supply.

Hence Power (P) =√3 × voltage (V) × current (I)

P = √3 × V × I

Given that voltage (V) = 460 V, current (I) = 17 A. Hence:

P = √3 × V × I = √3 × 460 × 17 = 13544.64 VA

But 1000 VA = 1 kVA. Hence:

P=13544.64\ VA*\frac{1\ kVA}{1000\ VA}=13.54\ kVA

8 0
3 years ago
1. Define weight
bogdanovich [222]

Answer:

  1. weight is the product of mass andgravitational acceleration expressed in newtons.
  2. w=mg
  3. w=mg ....w=50kg X 9.8m/s...w=490N
4 0
3 years ago
When blue light of wavelength 460 nm falls on a single slit, the first dark bands on either side of center are separated by 45.0
lyudmila [28]

Answer:

width of slit =1.23× 10⁻⁶ m  

Explanation:

we know the condition of diffraction minima,

d sin θ = n λ

λ = wavelength     θ = angle between the central maxima and 1st  minima

d = slit width

for first minima  n = 1

now,

d =\dfrac{n \lambda}{sin \theta}

\theta = \dfrac{45^0}{2} = 22.5^0

d =\dfrac{1\times 460 \times 10^{-9}}{sin 22.5^0}

d = 1228 × 10⁻⁹ m  = 1.228× 10⁻⁶ m

d =  1.23× 10⁻⁶ m

width of slit =1.23× 10⁻⁶ m  

6 0
3 years ago
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