Explanation:
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(a) <u>The</u><u> </u><u>segment</u><u> </u>A shows acceleration as velocity increases with the increase in time.
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(b) <u>The</u><u> </u><u>segment</u><u> </u>C shows the object is slowing down as the time increases in segment C, the velocity decreases and afterwards it comes to rest.
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(c) The velocity is segment B is <u>4</u><u>0</u><u>m</u><u>/</u><u>s</u><u>.</u> And in the diagram there is no change in velocity.
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(d) The acceleration of segment B is <u>zero</u><u>.</u> As there in no change in curve and it is moving with uniform velocity.
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Answer:
B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.
Explanation:
B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.
Answer:
40 V
Explanation:
I will assume that the resistors are
100 and 3900 and 1000 OHMS <=====(NOT W)
In series , the resistances add together 100 + 3900 + 1000 = 5000 ohms total
V = IR
I = V / R so the total current will be 200 v / 5000 ohms = .04 amps
this is the current through all of the resistors
so for the 1000 ohm resistor V = IR .04 (1000) = 40 V
Answer:
55.96kJ
Explanation:
Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether
Volume (v) = 200mL, density (d) = 0.7138g/mL
Mass = d × v = 0.7138 × 200 = 142.76g
Enthalpy of vaporization of diethyl ether = 29kJ/mol
MW of diethyl ether (C2H5)2O = 74g/mol
Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g
Energy = 142.76g × 0.392kJ/g = 55.96kJ
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