Answer:
= 85.89 ° C
Explanation:
The linear thermal expansion process is given by
ΔL = L α ΔT
For the three-dimensional case, the expression takes the form
ΔV = V β ΔT
Let's apply this equation to our case
ΔV / V = -0.507% = -0.507 10-2
ΔT = (ΔV / V) 1 /β
ΔT = -0.507 10⁻² 1 / 1.15 10⁻³
ΔT = -4.409
–T₀ = 4,409
= T₀ - 4,409
= 90.3-4409
= 85.89 ° C
Answer:
a) R `= 3.5 ohms
b) energy decipated = 560J
Explanation:
V = I . R
R = V / I
R `= 70 / 20
R `= 3.5 ohms
2)energy decipated = 1/2ij²
energy decipated = 1/2 x 2.8 x (20)²
energy decipated = 560J
A falling raindrop
Kinetic energy and potential energy are both applied when a body or object is falling.
Answer
A certain scale is not calibrated correctly,
Mass displayed as 0.75 kilogram less than its actual mass.
The scale of display will be equal to '1'
The correlation between actual mass and displayed mass is 1 because correlation is independent of change of origin and scale and correlation between any value and value minus 0.75 will be one.
So, the correct answer will be "1"
