Answer:
4 days
either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2
n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2
Explanation:
Answer:
<em>To reverse the direction of an electric current, we simply reverse the voltage either automatically with the help of some switching circuitry or manually by changing the voltage source terminals connection. </em>
Explanation:
For electric current to flow, there must be a potential difference, usually referred to as the voltage. The electric current flow is analogous to the flow of water under the action of a pump, through a series of pipe connections. The voltage is similar to the driving action of the pump, and current flows the same way water flows. The resistance due to drag on the pipe wall is equivalent to electric resistance. For current to flow in the reverse direction, the voltage or rather, the potential difference is changed, causing the current to flow in the opposite direction. This can be done by switching the terminals of the voltage source, or by automatic means. The automatic switching can be done with a transistor based circuitry.
Answer:

Explanation:
Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:




Ratio of kilomoles oxygen to kilomole nitrogen is:


It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.
The empirical formula for the compound is:

Answer:
because solar enegry is a perpetual source of enegry and by using it non renewable sources of enegry can be preserved and also prevents from enegry crisis.
hope its helpful to uh
Given data
Power (P) = 50 hp,
= 50 × 746, we know that 1 hp = 746 W.
= 37300 Watts (Watt = J/s)
Work = 6.40 ×10⁴ J
Power is defined as rate of doing work and the unit of power is<em> Watt.</em>
Mathematically,
Power = (Work / time) Watts
= 6.40 ×10⁴ / time
37300 W = 6.40 ×10⁴ J /time (Where time in seconds)
=> time = Work/Power
= 6.40 ×10⁴/37300
= <em>1.74 seconds </em>
<em> </em><em>Therefore , the engine need 1.74 seconds to do 6.40 6.40 ×10⁴ J of work </em>
<em> </em>