Answer:
Part a)
Part b)
Explanation:
Part a)
For force conditions of two blocks we will have
now from above equations we have
now we know that
now from above equation we have
Part b)
When heavier block is removed and F = 908 N is applied at the end of the string then we have
<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
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
Answer:
5.3Km/hr
Explanation:
Velocity=Displacement/Time
D=4km;T=0.75hr
V=4/0.75=5.33..
Friend #1 gets at least 2/5 of a pizza.
Friend #2 gets at least 2/5 .
Friend #3 gets at least 2/5 .
Friend #4 gets at least 2/5 .
Friend #5 gets at least 2/5 .
Friend #6 gets at least 2/5 .
Sum . . . . . . . . . at least 12/5 of a pizza.
Simplify . . . . . . at least 2.4 pizzas.
-- If pizzas can be bought by the half, they should order at least <em>2-1/2 pizzas.</em>
-- If only whole pizzas have to be ordered, then they should order at least <em>3 pizzas.</em>
