Answer:
Ohm, a german physicist, found experimentally that the current in a metal conductor is directly proportional to the potential difference across its ends
Explanation:
Answer:
perk
Explanation:
Perks are additional compensation not usually offered as wages
Answer: the thermal conductivity of the sample is 22.4 W/m . °C
Explanation:
We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.
ASSUMPTIONS
1. Steady operating conditions exist since the temperature readings do not change with time.
2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.
3. The apparatus possess thermal symmetry
ANALYSIS
The electrical power consumed by resistance heater and converted to heat is:
Wₐ = V<em>I</em> = ( 110 V ) ( 0.4 A ) = 44 W
Q = 1/2Wₐ = 1/2 ( 44 A )
Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so
A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²
Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be
Q = kA ( ΔT/L ) → k = QL / AΔT
k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )
k = 22.4 W/m . °C
Answer:critical stress= 20.23 MPa
Explanation:
Since there was an internal crack, we will divide the length of the internal crack by 2
Length of internal crack, a = 0.7mm,
Half length = 0.7mm/2= 0.35mm changing to meters becomes
0.35/ 1000= 0.35 x 10 ^-3m
The formulae for critical stress is calculated using
σC = (2Eγs /πa) ¹/₂
σC = critical stress=?
Given
E= Modulus of Elasticity= 225GPa =225 x 10 ^ 9 N/m²
γs= Specific surface energy = 1.0 J/m2 = 1.0 N/m
a= Half Length of crack=0.35 x 10 ^-3m
σC= (2 x 225 x 10 ^ 9 N/m² x 1.0 N/m /π x 0.35 x 10 ^-3m)¹/₂
=(4.5 x 10^11/π x 0.35 x 10 ^-3)¹/₂
=(4.0920 x10 ^14)¹/₂
σC=20.23 x10^6 N/m² = 20.23 MPa