The power delivered to the coil is 721.31 Watts
<h3>How to determine the power</h3>
Given that;
- Power of the heating coil = 500W
- Voltage = 110V
- Diameter of the Nichrome wire = 0. 500mm, radius = 0. 500/2 = 0. 00025m
But the formula for power is given a;
Power, P = V²/R
Then , R = V²/P
R = (110)²/ 500 = 24. 2Ω
To determine the length,
Length, L = RA/ ρ
L = 24. 2 × ( 0. 0025)² × 3. 142/ 10^-6
L = 4. 32m
We also have that;
Resistance, R = Ro( 1 + α ΔT)
R = 24. 2 ( 1 + 0. 0004 × 1180)
R = 35. 62Ω
Current, I = V R
Current, I = 110/24.2 = 4. 5 A
Power delivered = I²R = (4.5)² × 35. 62 = 721.31 Watts
Thus, the power delivered to the coil is 721.31 Watts
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Given :
Mass =5kg
T
1
=20
∘
C,T
2
=100
∘
C
ΔT=100−20=80
∘
C
Q=m×C×ΔT
where C= specific heat capacity of water
=4200J/(kgK)
Q=5×4200×80
=1680000 Joule.
=1680KJ
just analyze it in this way:
20cos30*=10( radical 3 )
20sin30*=10
Explanation:
We start by using the conservation law of energy:
or
Simplifying the above equation, we get
We can rewrite this as
Note that the expression inside the parenthesis is simply the acceleration due to gravity 
so we can write
where 
is the launch velocity.
