Answer = 1000N
F = ma
M = 500kg
a= 2 m/s/s
Solve for F
<span>Answer:
The moments of inertia are listed on p. 223, and a uniform cylinder through its center is:
I = 1/2mr2
so
I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2
Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration:
t = Ia
-1.20 Nm = (0.0120984 kgm2)a
a = -99.19 rad/s/s
Now we have a kinematics question to solve:
wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s
w = 0
a = -99.19 rad/s/s
Let's find the time first:
w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s
t = 10.558 s = 10.6 s
And the displacement (Angular)
Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form
s = (u+v)t/2
Which is
q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s
q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians
But the problem wanted revolutions, so let's change the units:
q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
Answer: a) 335.8 μm; b) keeping the same radius, the new has double potential, V=1340V so if teh radius is also double the potentail is the same (V=670V).
Explanation: In order to explain this problem we have to consider the potential given for sphere respect to infinity ( V=0) in the form:
V=k*Q/R the we have
R=k*Q/V= 9*10^9*25*10^-12/670=335.8 *10^-6 m
When two drop join to form a single drop (considering with the same radius) we have:
V=k*2Q/R
So the new V is double the original,
V=9*10^9*2*25*10^-12/335.8*10^-6=1340V
if the final single drop has a 2R of radius so
V=k*2Q/2R= 670 V
It has the same original potential.
To thaw a product requires it to undergo the process of
being subjected to latent heat of fusion. In this process, there is no change
in temperature however there is a change of phase from solid to liquid. What is
being thawed here actually is the frozen ice into liquid water which is in the
vegetables.
From the references, the latent heat of fusion of water
is:
Δ H = 333.55 kJ / kg
Therefore the amount of heat required to thaw the frozen
vegetable is:
Heat required = Δ H * m
Heat required = (333.55 kJ / kg) (0.450 kg)
Heat required = 150.10 kJ
In 1 kcal there is 4.184 kJ, therefore:
Heat required = 150.10 kJ (1 kcal / 4.184 kJ)
<span>Heat required = 35.87 kcal</span>