Answer:
Power_input = 85.71 [W]
Explanation:
To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.
where:
W = work [J] (units of Joules)
F = force [N] (units of Newton)
d = distance [m]
We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.
Now replacing:
![W = (80*10)*3\\W = 2400 [J]](https://tex.z-dn.net/?f=W%20%3D%20%2880%2A10%29%2A3%5C%5CW%20%3D%202400%20%5BJ%5D)
Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.
where:
P = power [W] (units of watts)
W = work [J]
t = time = 40 [s]
![P = 2400/40\\P = 60 [W]](https://tex.z-dn.net/?f=P%20%3D%202400%2F40%5C%5CP%20%3D%2060%20%5BW%5D)
The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.
![Effic=0.7\\Effic=P_{required}/P_{introduced}\\P_{introduced}=60/0.7\\P_{introduced}=85.71[W]](https://tex.z-dn.net/?f=Effic%3D0.7%5C%5CEffic%3DP_%7Brequired%7D%2FP_%7Bintroduced%7D%5C%5CP_%7Bintroduced%7D%3D60%2F0.7%5C%5CP_%7Bintroduced%7D%3D85.71%5BW%5D)
Answer:
<h3>The answer is 45 J</h3>
Explanation:
The work done by an object can be found by using the formula
<h3>workdone = force × distance</h3>
From the question
distance = 3 meters
force = 15 newtons
We have
workdone = 15 × 3
We have the final answer as
<h3>45 J</h3>
Hope this helps you
So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.
<h3>Power radiated by the radiant wall heater</h3>
The power radiated by the radiant wall heater is given by P = εσAT⁴ where
- ε = emissivity = 1 (since we are not given),
- σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
- A = surface area of cylindrical wall heater = 2πrh where
- r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
- h = length of heater = 0.6 m, and
- T = temperature of heater
Since P = εσAT⁴
P = εσ(2πrh)T⁴
Making T subject of the formula, we have
<h3>Temperature of heater</h3>
T = ⁴√[P/εσ(2πrh)]
Since P = 1.5 kW = 1.5 × 10³ W
Substituting the values of the variables into the equation, we have
T = ⁴√[P/εσ(2πrh)]
T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]
T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]
T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]
T = ⁴√[0.01105 × 10¹⁴ K⁴)]
T = ⁴√[1.105 × 10¹² K⁴)]
T = 1.0253 × 10³ K
T = 1025.3 K
So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
Learn more about temperature of radiant wall heater here:
brainly.com/question/14548124
Answer:
radiation is the correct answer
Answer:
The index of refraction of the liquid is 1.35.
Explanation:
It is given that,
Critical angle for a certain air-liquid surface, 
Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1
Using Snell's law for air liquid interface as :
So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.
