<span>here's a cheap trick
it would take the same time to accelerate from rest to top speed
as it would take to decelerate from top speed to zero
so
instead of
d = Vi t + 1/2 a t^2 where Vi is positive and a is negative
we'll use
Vi = 0 and a is positive
giving
85 = 0 + 1/2 (0.43) t^2 = 0.215 t^2
t^2 = 395.345
t = 19.88s or 20. s to 2 sig figs
or we ccould find Vi from
Vf*2 = Vi^2 + 2 a d
0 = Vi^2 + 2 (0.43) 85
Vi^2 = 71.4
Vi = 8.45m/s
then
85 = 8.45 t + 1/2 (-0.43) t^2
85 = 8.45 t - 0.215 t^2
0.215 t^2 - 8.45 + 85 = 0
t = 19.65s or 20. s to 2 s.f.(minor difference arises from rounding Vi)
or another cheap trick
when a is constant
Vavg = (Vf + Vi) /2 = 8.45/2 = 4.225
and
d = Vavg t
85 = 4.225 t
t = 20.12 or 20. s to 2 s.f. (minor differences from intermidiate roundings)
anyway you choose you get 20. s</span>
Answer: 70 N to the East
Explanation:
lets assume east is positive and west is negative, since they are in opposite directions the net external force = F1+F2
Net force = (-60) + 130
Net force = 70
or
Net force = 70 N in the east direction
Answer:
In a controlled experiment, an independent variable (the cause) is systematically manipulated and the dependent variable (the effect) is measured; any extraneous variables are controlled. The researcher can operationalize (i.e. define) the variables being studied so they can be objectivity measured.
Answer:
H vaporization = 100.0788 kJ/mol
Explanation:
Use clausius clapyron's adaptation for the calculation of Hvap as:
Where,
P₂ and P₁ are the pressure at Temperature T₂ and T₁ respectively.
R is the gas constant.
T₂ = 823°C
T₁ = 633°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So, the temperature,
T₂ = (823 + 273.15) K = 1096.15 K
T₁ = (633 + 273.15) K = 906.15 K
P₂ = 400.0 torr , P₁ = 40.0 torr
R = 8.314 J/K.mol
Applying in the formula to calculate heat of vaporization as:
Solving for heat of vaporization, we get:
H vaporization = 100078.823 J/mol
Also, 1 J = 10⁻³ kJ
So,
<u>H vaporization = 100.0788 kJ/mol</u>
