Given:
Water, 2 kilograms
T1 = 20 degrees Celsius, T2 = 100
degrees Celsius.
Required:
Heat produced
Solution:
Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2 kilograms (4.184 kiloJoules
per kilogram Celsius) (100 degrees Celsius – 20 degrees Celsius)
<u>Q (heat) = 669.42 Joules
</u>This is the amount of heat
produced in boiling 2 kg of water.
<span>B.Extrinsic motivation </span>
Answer:
Earth's water is always in movement, and the natural water cycle, also known as the hydrologic cycle, describes the continuous movement of water on, above, and below the surface of the Earth. Water is always changing states between liquid, vapor, and ice, with these processes happening in the blink of an eye and over millions of years.
Hope this helped!! :))
Explanation:
S= 343m/s
F=256Hz
WL= 343ms/256-1
WL=V/F
= 1.339844m
Answer:
The radius of coil 2 = 2.7 cm
Explanation:
The number of coils = 2
It is given that both carry equal current and rotates in the magnetic field.
The given radius of coil 1 = 4.0 cm
Coil 1 rotates = 0.21 T field
Coil 2 rotates = 0.45 T filed.
The radius of coil 2 need to be calculated.
Torque action on dipole is given by
here T1 = T2
