<span>A+B-C
</span><span>A = 6x - 2y
B = -4x - 8y
C = -3x + 9y
(</span>6x - 2y) + (-4x - 8y) - (-3x + 9y)
(6x - 2y) + (-4x - 8y) + (3x - 9y)
2x -10y + (3x - 9y)
5x - 19y
Answer:
Temperature increase = 2.1 [C]
Explanation:
We need to identify the initial data of the problem.
v = velocity of the copper sphere = 40 [m/s]
Cp = heat capacity = 387 [J/kg*C]
The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be:
![E_{k}=Q\\ E_{k}= kinetic energy [J]\\Q=thermal energy [J]\\Re-employment values and equalizing equations\\\\\frac{1}{2} *m*v^{2}=m*C_{p}*dT \\The masses are canceled \\\\dT=\frac{v^{2}}{C_{p} *2} \\dT=2.1 [C]](https://tex.z-dn.net/?f=E_%7Bk%7D%3DQ%5C%5C%20E_%7Bk%7D%3D%20kinetic%20energy%20%5BJ%5D%5C%5CQ%3Dthermal%20energy%20%5BJ%5D%5C%5CRe-employment%20values%20and%20equalizing%20equations%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%3Dm%2AC_%7Bp%7D%2AdT%20%20%5C%5CThe%20masses%20are%20canceled%20%5C%5C%5C%5CdT%3D%5Cfrac%7Bv%5E%7B2%7D%7D%7BC_%7Bp%7D%20%2A2%7D%20%5C%5CdT%3D2.1%20%5BC%5D)
Answer:
1.08 m/s
Explanation:
This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.
Time taken to fall 9.5 m
vertical acceleration = a = 9.8 m/s^2.
vertical velocity = 0, (since there is only horizontal component for velocity,
)
distance traveled s = 9.5 m.
Substituting these values in the equation



⇒ t= 1.392 sec
Velocity needed
We know the time taken (1.392 s) to travel 1.5 m,
So velocity = 1.5 m / 1.392 s = 1.08 m/s
hence velocity of the diver must be at least 1.08 m/s
Do 25-14 and you will get your answer
Answer:
a)
Weight in Air = 0.3N
Weight in Water = 0.25N
Weight in Liquid = 0.24N.
Upthrust /Buoyant Force = Weight in Air – Weight in Fluid(Water in this case)
= 0.3 – 0.25
= 0.5N.
b) R.D of Body = Density of Body/Density of Standard Fluid(Water).
There's a Derived Formula for RD.
I'm gonna Apply it here.
Ask me for the derivation in the Comment section if you need it.
RD = α/ρ = (Weight in Air) / (Upthrust Force)
Where
α = density of the Body(or reference substance)
ρ = density of standard fluid (water)
= 0.3/0.05 = 6.
c) RD of Liquid = (Density of Liquid) /(Density of standard Fluid(water)
Or we just go by that formula
RD of Liquid = Weight in Air/Upthrust(In Liquid)
We'll be using the Upthrust in that Liquid now.
= 0.3 – 0.24 = 0.06
RD = 0.3/0.06 = 5.