Answer:
Power, P = 722.96 watts
Explanation:
It is given that,
Voltage, V = 120 V
Length of nichrome wire, l = 8.9 m
Diameter of wire, d = 0.86 mm
Radius of wire, r = 0.43 mm = 0.00043 m
Resistivity of wire, 
We need to find the power drawn by this heater. Power is given by :

And, 


P = 722.96 watts
So, the power drawn by this heater element is 722.96 watts. Hence, this is the required solution.
As per given conditions there are two directions along which forces are acting
1. Net force along left direction is given as

2. Net force towards right direction is given as

now since the two forces here in opposite direction so here we will have net force given as



so here net forces must be 440 N towards right
Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:
diameter = 40 mm ==> area = <em>π</em> (0.04 m)² ≈ 0.00503 m²
diameter = 120 mm ==> area = <em>π</em> (0.12 m)² ≈ 0.0452 m²
Then the speed at the end of the 40 mm pipe is
(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s
(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s
Answer:
Explanation:
To find the half-lifes of the isotope we need to use the following equation:
(1)
<em>where Nt: is the amount of the isotope that has not yet decayed after a time t, N₀: is the initial amount of the isotope, t: is the time and </em>
<em>: is the half-lifes.</em>
By solving equation (1) for t we have:
<u>Having that:</u>
Nt = 450
N₀ = 3150 + 450 = 3600,
The half-lifes of the isotope is:

Therefore, 3 half-lives of the isotope passed since the rock was formed.
I hope it helps you!
Answer: SG = 2.67
Specific gravity of the sand is 2.67
Explanation:
Specific gravity = density of material/density of water
Given;
Mass of sand m = 100g
Volume of sand = volume of water displaced
Vs = 537.5cm^3 - 500 cm^3
Vs = 37.5cm^3
Density of sand = m/Vs = 100g/37.5 cm^3
Ds = 2.67g/cm^3
Density of water Dw = 1.00 g/cm^3
Therefore, the specific gravity of sand is
SG = Ds/Dw
SG = (2.67g/cm^3)/(1.00g/cm^3)
SG = 2.67
Specific gravity of the sand is 2.67