Answer:
acceleration a = 1.04 m/s2
Explanation:
Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:
s = 75 + 180 = 255 m
We can use the following equation of motion to find out the distance traveled by the car:
where v = 23 m/s is the velocity of the car when it passes the worker, 
= 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.
<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’):
</em></h2><h2><em>
</em></h2><h2><em>The car’s speed is 25m/s
</em></h2><h2><em>The distance travelled is 75m
</em></h2><h2><em>Then we have the formulas for speed and distance:
</em></h2><h2><em>
</em></h2><h2><em>v = a x t -> 25 = a x t
</em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2
</em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say:
</em></h2><h2><em>
</em></h2><h2><em>t = 25 / a
</em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a
</em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1):
</em></h2><h2><em>
</em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2
</em></h2><h2><em>t^2 = 150 / a
</em></h2><h2><em>Since t has the same value for both truths we can say:
</em></h2><h2><em>
</em></h2><h2><em>625 / a^2 = 150 / a
</em></h2><h2><em>
</em></h2><h2><em>Thus multiply both sides with a^2:
</em></h2><h2><em>
</em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17
</em></h2><h2><em>
</em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>
D is the correct answer!!
Answer:
1.97×10⁻²¹ J
Explanation:
Use ideal gas law to find temperature.
PV = nRT
(9 atm) (9 L) = (83.3 mol) (0.0821 L·atm/mol/K) T
T = 11.9 K
The average kinetic energy per atom is:
KE = 3/2 kT
KE = 3/2 (1.38×10⁻²³ J/K) (11.9 K)
KE = 2.46×10⁻²² J
For a mass of 5.34×10⁻²⁶ kg, the kinetic energy is:
KE = (5.34×10⁻²⁶ kg) (1 mol / 0.004 kg) (6.02×10²³ atom/mol) (2.46×10⁻²² J)
KE = 1.97×10⁻²¹ J
