2.39 Watts roughly since watts is joules per second it’s just 910j/380s
Lithium metals reacts slowly with water to form a colourless solution of lithiums hydroxide (LiOH) and hydrogen gas (H2). The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic, but the reaction is slower than that of sodium (immediately below lithium in the periodic table)
Answer:
El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.
Explanation:
El gasto es el flujo volumétrico de gasolina (
), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:
(1)
Donde:
- Diámetro de la manguera, medido en pies.
- Velocidad medida de salida, medida en pies por segundo.
Si sabemos que
y
, entonces el gasto de gasolina es:


El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.
Answer:
Cp= 0.44 J/g.C
This is heat capacity of metal.
Explanation:
From energy conservation
Heat lost by metal = Heat gain by water +Heat gain by calorimeter
Because here temperature of metal is high that is why it loose the heat.The temperature of water and calorimeter is low that is why they gain the heat.
final temperature is T= 30.5 C
We know that sensible heat transfer given as
Q= m Cp ΔT
m=Mass
Cp=Specific heat capacity
ΔT=Temperature difference
By putting the values
55 x Cp ( 99.5 - 30.5) = 40 x 4.184 ( 30.5- 21 ) + 10 x ( 30.5 - 21)
Cp ( 99 .5- 30.5) = 30.65
Cp= 0.44 J/g.C
This is heat capacity of metal.
Answer:
The electric field is 
Explanation:
Given that,
Radius = 2.00 cm
Number of turns per unit length 
Current 
We need to calculate the induced emf

Where, n = number of turns per unit length
A = area of cross section
=rate of current
Formula of electric field is defined as,

Where, r = radius
Put the value of emf in equation (I)
....(II)
We need to calculate the rate of current
....(III)
On differentiating equation (III)

Now, put the value of rate of current in equation (II)


Hence, The electric field is 