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.
Answer:

Explanation:
Given:
- mass of car,

- distance of skidding after the application of brakes,

- coefficient of kinetic friction,

<u>So, the energy dissipated during the skidding of car:</u>
<em>Frictional force:</em>

where N = normal reaction by ground on the car


<em>Now from the work-energy equivalence:</em>


is the dissipated energy.
Answer:
6858.5712 m/s
Explanation:
Given that:
Radius, r
R = 3.20 * 10^3.
Normal force = 0.5 * normal weight
Normal force = Fn ; Normal weight = Fg
Fn = 0.5Fg
Recall:
mv² / R = Fn + Fg
Fn = 0.5Fg
mv² / R = 0.5Fg + Fg
mv² /R = 1.5Fg
mv² = 1.5Fg * R
F = mg
mv² = 1.5* mg * R
v² = 1.5gR
v = sqrt(1.5gR)
V = sqrt(1.5 * 9.8 * 3.2 * 10^3)
V = sqrt(47.04^3)
V = 6858.5712 m/s
(1) Changing Fahrenheit to Celsius:
The formula used to convert from Fahrenheit to Celsius is as follows:
C = <span>(F - 32) * 5/9
</span>We are given that F=200, substitute in the above formula to get the corresponding temperature in Celsius as follows:
C = (200-32) * (5/9) = 93.333334 degrees Celsius
(2) Changing the Fahrenheit to kelvin:
The formula used to convert from Fahrenheit to kelvin is as follows:
K = <span>(F - 32) * 5/9 + 273.15
</span>We are given that F = 200. substitute in the above formula to get the corresponding temperature in kelvin as follows:
K = (200-32)*(5/9) + 273.15 = 366.483334 degrees kelvin
Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms
Explanation: In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:
I(t)=io*Exp(-t/τ)
and also we consider that io=V/R=(1.5/6.43*10^3)
=233.28 A
then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6
=22.31 ms
Finally the time to reduce the current to 2.57% of its initial value is obtained from:
I(t)=io*Exp(-t/τ) for I(t)/io=0.0257=Exp(-t/τ) then
ln(0.0257)*τ =-t
t=-ln(0.0257)*τ=81.68 ms