Answer:
Ф = 239.73 rad
Explanation:
α = 12 + 15×t
W = ∫α×dt
= ∫(12 + 5×t)×dt
= 12×t + 2.5×t^2
then:
Ф = ∫W×dt
= ∫(12×t + 2.5×t^2)dt
= 6×t^2 + 5/6×t^3
therefore the angle at t = 4.88s is:
Ф = 6×(4.88)^2 + 5/6×(4.88)^3
= 239.73 rad
Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
The right answer is
all of the above
good luck
Answer: 2.83 J/mol
Explanation:
Heat of solution, sometimes interchangeably called enthalpy of solution, is said to be the energy released or absorbed when the solute dissolves in the solvent. A solute is that which can dissolve in a solvent, to form a solution
Given
No of moles of CaCl = 7.5 mol
Total energy used = 21.2 J
Heat of solution = q/n where
q = total energy
n = number of moles
Heat of solution = 21.2 / 7.5
Heat of solution = 2.83 J/mol
<u>Answer:</u>
<em>The correct equation for measuring the average microscopic weight for 3 isotopes is multiply the rate of abundance by each weight and add them.</em>
<u>Explanation:</u>
To calculate the average microscopic mass of element using weights and relative abundance we have to follow the following steps.
- Take the correct weight of each isotope (that will be in decimal form)
- Multiply the weight of each isotope by its abundance
- Add each of the results together.
<em>This gives the required average microscopic weight of the three isotopes.</em>
