Answer:
The maximum speed will be 26.475 m/sec
Explanation:
We have given mass of the toy m = 0.50 kg
radius of the light string r = 1 m
Tension on the string T = 350 N
We have to find the maximum speed without breaking the string
For without breaking the string tension must be equal to the centripetal force
So 
So 
v = 26.475 m /sec
So the maximum speed will be 26.475 m/sec
Answer:
1500 per second.
Explanation:
vibrations = 1.5 kilohertz
1.5×1000=1500
the answer is 1500 per second.
A heat engine is a device that uses heat to produce useful work.
When you bring two objects of different temperature together, energy will always be transferred from the hotter to the cooler object. The objects will exchange thermal energy, until thermal equilibrium is reached, i.e. until their temperatures are equal. We say that heat flows from the hotter to the cooler object. Heat is energy on the move.
Units of heat are units of energy. The SI unit of energy is Joule. Other often encountered units of energy are 1 Cal = 1 kcal = 4186 J, 1 cal = 4.186 J, 1 Btu = 1054 J.
Without an external agent doing work, heat will always flow from a hotter to a cooler object. Two objects of different temperature always interact. There are three different ways for heat to flow from one object to another. They are conduction, convection, and radiation.
Answer:
The heat loss per unit length is 
Explanation:
From the question we are told that
The outer diameter of the pipe is 
The thickness is 
The temperature of water is 
The outside air temperature is 
The water side heat transfer coefficient is 
The heat transfer coefficient is 
The heat lost per unit length is mathematically represented as
![\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%5Cpi%20%28T%20-%20Ta%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_1%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_2%7D%7D)
Substituting values
![\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%2A%203.142%20%28363%20-%20263%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B300%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B20%7D%7D)
