Answer:
a = 1,008 10⁻³ m / s²
Explanation:
For this exercise, let's use the kinematic relations of accelerated motion
v² = v₀² - 2 a x
The negative sign is because the acceleration is opposite to the speed, the final speed is zero
0 = v₀² - 2 a x
a = v₀² / 2x
Let's reduce the magnitudes to the SI system
x = 2.4mm (1m / 10³mm) = 2.4 10⁻³m
Let's calculate
a = 2.2²/2 2.4 10⁻³
a = 1,008 10⁻³ m / s²
It is to be noted that to answer this question, we only use the given mass and the given amount of heat needed to liquefy the object. The equation that we can use is,
Hf = mhf
where Hf is the amount of heat, m is the mass, and hf is the heat of fusion.
59400 J = (0.63 kg)hf
The value of hf from the equation is <em>94,285.71 J/kg</em>.
Answer:
Given R (t) = 60/(1+t²), ft
/min
To find the amount of air in ft³, during the first min,
R (t) = 60/(1+t²
at t=1min, R is the air amount in ft³/min
Take the integral, and evaluate over [0,1]
Integral Of R (t) = 60/(1+t²) = 60 tan⁻¹(t),
60(Tan⁻¹(1) - Tan⁻¹(0)) = 60(pi/2) = 30 π
Therefore, it means that in the first minute, 30π ft³ of air escaped.
Answer:
PLEASE HELP John is rollerblading down a long, straight path. At time zero, there is a mailbox about 1 m in front of him. In the 5 s time period that follows, John's velocity is given by the velocity versus time graph in the figure. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph. John is rollerblading down a long, straight path. At time zero, there is a mailbox about 1 m in front of him. In the 5 s time period that follows, John's velocity is given by the velocity versus time graph in the figure. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph. Assuming that all the numbers given are exact, what is John's position at a time of 4.79 s ? Enter your answer to at least three significant digits.
I would say the greatest amount of error is introduced by eyeballing the flask at two different levels. You're supposed to measure liquids at eye level. The angle viewing downward could be off by a good amount
