What is an example of how you can use scientific inquiry to solve a real life problem.
Answer:
Explanation:
We need the power equation for this which is
P = Work/time
We have everything we need to solve this (the mass of the object is extra information):
P = 6860/4
P = 1715W
Answer:
m = 35.98 Kg ≈ 36 Kg
Explanation:
I₀ = 125 kg·m²
R₁ = 1.50 m
ωi = 0.600 rad/s
R₂ = 0.905 m
ωf = 0.800 rad/s
m = ?
We can apply The law of conservation of angular momentum as follows:
Linitial = Lfinal
⇒ Ii*ωi = If*ωf <em>(I)</em>
where
Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m
If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m
Now, we using the equation <em>(I) </em>we have
(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800
⇒ m = 35.98 Kg ≈ 36 Kg
<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q</h2>
Explanation:
Specific heat capacity
It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .
It is given as :
Heat absorbed = mass of substance x specific heat capacity x rise in temperature
or ,
Q= m x c x t
In above question , it is given :
For Q
mass of Q = m
Temperature changed =T₂/2
Heat supplied = x
Q= mc t
or
X=m x C₁ X T₁
or, X =m x C₁ x T₂/2
or, C₁=X x 2 /m x T₂ (equation 1 )
For another quantity : P
mass of P =m/2
Temperature= T₂
Heat supplied is same that is : X
so, X= m/2 x C₂ x T₂
or, C₂=2X/m. T₂ (equation 2 )
Now taking ratio of C₂ to c₁, We have
C₂/C₁= 2X /m.T₂ /2X /m.T₂
so, C₂/C₁= 1/1
so, the ratio is 1: 1