If iodine is added to a starch solution, they react with each other and the iodine darkens to an almost pitch black.
however, if iodine is added to a solution containing no starch, it will show up only as an extremely pale brown. almost colorless and hardly visible.
when following the changes in some inorganic oxidation reduction reactions, iodine may be used as an indicator to follow the changes of iodide ion and iodine element. soluble starch solution is added. only iodine element in the presence of iodide ion will give the characteristic blue black color. neither iodine element alone nor iodide ions alone will give the color result.
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The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
As per above given data
initial velocity = 19.3 km/s
final velocity = - 18.8 km/s
now in order to find the change in velocity
Part b)
Now we need to find acceleration
acceleration is given by formula
given that
now the acceleration is given as
so above is the acceleration
Answer:
143.352 watt.
Explanation:
So, in the question above we are given the following parameters or data or information that is going to assist us in answering the question above efficiently. The parameters are:
"A 1.8 m wide by 1.0 m tall by 0.65m deep home freezer is insulated with 5.0cm thick Styrofoam insulation"
The inside temperature of the freezer = -20°C.
Thickness = 5.0cm = 5.0 × 10^-2 m.
Step one: Calculate the surface area of the freezer. That can be done by using the formula below:
Area = 2[ ( Length × breadth) + (breadth × height) + (length × height) ].
Area = 2[ (1.8 × 0.65) + (0.65 × 1.0) + (1.8 × 1.0)].
Area = 7.24 m^2.
Step two: Calculate the rate of heat transfer by using the formula below;
Rate of heat transfer =[ thermal conductivity × Area (T1 - T2) ]/ thickness.
Rate of heat transfer = 0.022 × 7.24(25+20)/5.0 × 10^-2 = 143.352 watt.
Answer:
v’= 9.74 m / s
Explanation:
The Doppler effect is due to the relative movement of the sound source and the receiver of the sound, in this case we must perform the exercise in two steps, the first to find the frequency that the bat hears and then the frequency that the audience hears that also It is sitting.
Frequency shift heard by the murciela, in case the source is still and the observer (bat) moves closer
f₁ ’= f₀ (v + v₀)/v
Frequency shift emitted by the speaker in the bat, in this case the source is moving away from the observer (public sitting) that is at rest
f₂’= f₁’ v/(v - vs)
Note that in this case the bat is observant in one case and emitter in the other, called its velocity v’
v’= vo = vs
Let's replace
f₂’= f₀ (v + v’)/v v/(v -v ’)
f₂’= f₀ (v + v’) / (v -v ’)
(v –v’ ) f₂’ / f₀ = v + v ’
v’ (1+ f₂’ /f₀) = v (f₂’/fo - 1)
v’ (1 + 1.059) = 340 (1.059 - 1)
v’= 20.06 / 2.059
v’= 9.74 m / s
