<h3><u>Answer;</u></h3>
<em><u> = 48,828.125 mi/hr²</u></em>
<h3><u>Explanation and solution</u>;</h3>
- <em><u>Centripetal acceleration is the rate of change of angular velocity. Centripetal acceleration occurs towards the center of the circular path along the radius of the circular path</u></em>.
- Centripetal acceleration is given by; <em>V²/r ; </em>
<em>V = 125 mi/h and r = 0.320 miles </em>
- <em>Thus; centripetal acceleration = 125²/0.320 </em>
=15625/0.320
<em><u> = 48,828.125 mi/hr²</u></em>
Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.
We have the following equation representing the half-life decay:
A is the resulting amount after t time
Ao is the initial amount = 50 mg
t= Elapsed time
t half is the half-life of the substance = 14.3 days
We replace the know values into the equation to have an exponential decay function for a 50mg sample
That would be the answer for a)
To know the P-32 remaining after 84 days we have to replace this value in the equation:
So, after 84 days the P-32 remaining will be 0.85 mg
The pressure of gas will increase because gaseous state is the final state and even if the heat added is evaporating some more gas is still added. It also depends on the temperature of heat added, if the temperature doesn't change the it's most likely for the pressure to be stable...
Hope it helps
The molecular formula =C₆H₁₂O₆
<h3>Further explanation</h3>
Given
6.00 g of a certain compound X
The molecular molar mass of 180. g/mol
CO₂=8.8 g
H₂O=3.6 g
Required
The molecular formula
Solution
mass C in CO₂ :
= 1.12/44 x 8.8
= 2.4 g
mass H in H₂O :
= 2.1/18 x 3.6
= 0.4 g
Mass O in compound :
= 6-(2.4+0.4)
= 3.2 g
Mol ratio C : H : O
= 2.4/12 : 0.4/1 : 3.2/16
= 0.2 : 0.4 : 0.2
= 1 : 2 : 1
The empirical formula : CH₂O
(CH₂O)n=180 g/mol
(12+2+16)n=180
(30)n=180
n=6
(CH₂O)₆=C₆H₁₂O₆
Answer:
it's option c
Explanation:
because if I'm not wrong I have learned these type of questions back 11 and I remember that rutherfords observation was few alpha particles were deflected by small angles.