Answer: The unit of impulse is applied to an object produces an equivalent vector change in its linear momentum, also in the same direction.
Explanation:
Answer:
Yes the body will receive a dangerous shock in both cases.
Explanation:
Different parts of the body has different resistance. skin has the high resistance as compared to other organs of the body.
Dry skin has high resistance than wet skin this is because water is relatively good conductor of electricity, it adds parallel path to the current flow and hence reduces skin resistance.
Dry hands body has approximately 500 kΩ resistance and if 120 V electricity supply current received will be:
I = V/R= 120/ 500*10^3
I= 0.24 mA
Even the current seems is much lower than the safe zone but this is the case in case of DC voltage in case of AC voltage the body will receive a shock this is because the skin pass more current when the voltage is changing i.e. AC.
Similarly for wet hands body resistance is 1 kΩ. so the current through the body seems to be:
I = 120 / 1000
I = 12 mA
The current is higher than safe zone so the body will receive a dangerous shock.
The cost of developing thermonuclear power with plasmabe defended because D. It can provide an inexpensive power source.
<h3>How did the
cost of developing t
hermonuclear power defended?</h3>
The cost of developing thermonuclear power defended becvause we can see in the paragraph how it was told that the generation of ths power can be donee through the understanding of the occurrence of plasmain nature,
It should be noted that this thermonuclear power with plasmabe posses the characteristics which make it to exist in the ionosphere, and it can be felt in the flames as well; as in the chemical and nuclearexplosions.
In conclusion the power can be seen as an inexpensive source power because the p[roduction of this power cn be found in most of the thing that can be found around us as discused above.
Therefore, option D is correct.
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If the primary wire's power is 10 A and one branch's power is 4 A, another branch's power will be 6A.
According to Kirchhoff's current law (KCL), the total current flowing through a parallel route circuit's junction equals the total current flowing away from it.
Provided that one of the two branches through which power exits the intersection has a flow of 4A, and also that the junction's overall flow entering it is 10A, the entire current going the junction should be 10A.
Consequently, the second wire's power may be expressed as;
I = I1+ I2 [ where I= total current (10A);
I1= current in one branch (4A) &
I2= current in another branch]
⇒I2 = I - I1
⇒I2 = 10A - 4A
⇒I2 = 6A
Therefore, it can be concluded that when the primary wire bears 10A power having 4A in one of its branches, another branch carries 6A power.
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A reference is critical because it determines how much motion has occurred. You can't determine how much an objected has moved if you don't know where it came from.
