Answer:
Explanation:
If we let our reference frame travel at 30 m/s with the constant speed car, The accelerating car increases its velocity by 10 m/s in 3 seconds.
The average velocity of the accelerating car is (0 + 10) / 2 = 5 m/s
It will advance its position 5 m/s(3 s) = 15 m in the accelerating period.
It takes 5 + 3 = 8 m for the two cars to become side by side.
It would take another 5 + 3 = 8 m for the accelerating car to leave a gap of 3 m between.
The car requires 8 + 8 = 16 m to pass the other safely but the acceleration period only gets him to 15 m.
So despite your saying this is not a YES / NO question, the answer is NO the acceleration is too low or not long enough to meet the required clearances.
Input needed is 10000 J/s / 0.30 = 333333 = J/s
three hours requires 333333(3)(3600) = 360 MJ of energy
360 MJ / 34 MJ/liter = 10.6 liters.
D. 128.25 because force=mass x acceleration
Temperature is just a measure of how HOT or COLD a substance is, which can be easily defined by a magnitude using a numerical value say “300 K” or “27°C”. Hence we can say it is a scalar quantity.
But the energy which transfer by virtue of a temperature difference is a vector quantity, as it has both magnitude and direction of motion (from High temperature to low temperature region).
Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles
