Answer:
the glucose in air will solidify and it will stop the process.
it will cause hot air
Answer:
Specific heat at constant pressure is = 1.005 kJ/kg.K
Specific heat at constant volume is = 0.718 kJ/kg.K
Explanation:
given data
temperature T1 = 50°C
temperature T2 = 80°C
solution
we know energy require to heat the air is express as
for constant pressure and volume
Q = m × c × ΔT ........................1
here m is mass of the gas and c is specific heat of the gas and Δ
T is change in temperature of the gas
here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.
and here at constant pressure Specific heat is greater than the specific heat at constant volume,
so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is
Specific heat at constant pressure is = 1.005 kJ/kg.K
and
Specific heat at constant volume is = 0.718 kJ/kg.K
Answer:
yes
Explanation:
this is simple
the horizontal line is adjacent
the vertical line is opposite
recall that cos x=adj/hyp
adj=hyp(cos x)
while opp=hyp(sin x)
Answer:
240 Ω
Explanation:
Resistance: This can be defined as the opposition to the flow of current in an electric field. The S.I unit of resistance is ohms (Ω).
The expression for resistance power and voltage is give as,
P = V²/R.......................... Equation 1
Where P = Power, V = Voltage, R = Resistance
Making R the subject of the equation,
R = V²/P.................... Equation 2
Given: V = 120 V, P = 60 W.
Substitute into equation 2
R = 120²/60
R = 240 Ω
Hence the resistance of the bulb = 240 Ω
