We can substitute the given values into the equation for T, given the surrounding temperature T0 = 0, initial temperature T1 = 140, constant k = -0.0815, and time t = 15 minutes.
T = 0 + (140 - 0)e^(-0.0815*15) = 140e^(-1.2225) = 41.23°F
By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
Answer:
0.08 sin 3nt +
metre. Then calculate-
a time period
(d) Time period
(
Minitial phase
wõisplacement
7
(c) Displacement from mean position at t=
36
sec.
Explanation:
thats the answer
Answer:
1.3 x 10^(-2) atm/s
Explanation:
It follows the stoichiometry. For every mole of O3 that disappears, 1.5 moles (that is, 3/2) of O2 appears:
1.5 * 0.009 atm/s = 0.0135 atm/sec; the answer is 1.3 x 10^(-2) atm/s
Answer:
A theory is a system of ideas intended to explain something, and a hypothesis is an educated guess.
Explanation: Hope this Helps! :)
