Answer:
200 km/hr
Explanation:
Since he goes 80km per hour, multiply this by 2.5 or two and a half hours.
80 x 2.5 = 200 km/hr.
Answer:
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Answer:
20.62361 rad/s
489.81804 J
Explanation:
= Initial moment of inertia = 9.3 kgm²
= Final moment of inertia = 5.1 kgm²
= Initial angular speed = 1.8 rev/s
= Final angular speed
As the angular momentum of the system is conserved
![I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s](https://tex.z-dn.net/?f=I_i%5Comega_i%3DI_f%5Comega_f%5C%5C%5CRightarrow%20%5Comega_f%3D%5Cdfrac%7BI_i%5Comega_i%7D%7BI_f%7D%5C%5C%5CRightarrow%20%5Comega_f%3D%5Cdfrac%7B9.3%5Ctimes%201.8%7D%7B5.1%7D%5C%5C%5CRightarrow%20%5Comega_f%3D3.28235%5C%20rev%2Fs%3D3.28235%5Ctimes%202%5Cpi%3D20.62361%5C%20rad%2Fs)
The resulting angular speed of the platform is 20.62361 rad/s
Change in kinetic energy is given by
![\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J](https://tex.z-dn.net/?f=%5CDelta%20K%3D%5Cdfrac%7B1%7D%7B2%7D%28I_f%5Comega_f%5E2-I_i%5Comega_i%5E2%29%5C%5C%5CRightarrow%20%5CDelta%20K%3D%5Cdfrac%7B1%7D%7B2%7D%285.1%5Ctimes%20%2820.62361%29%5E2-9.3%5Ctimes%20%281.8%5Ctimes%202%5Cpi%29%5E2%29%5C%5C%5CRightarrow%20%5CDelta%20K%3D489.81804%5C%20J)
The change in kinetic energy of the system is 489.81804 J
As the work was done to move the weight in there was an increase in kinetic energy
Answer:
distance = 33.124 meters
Explanation:
To solve this question, we will use one of the equations of motion which is:
s = ut + 0.5a * t^2
where:
s is the distance that we want to get
u is the initial velocity = 0
a is the acceleration due to gravity = 9.8 m/sec^2
t is the time = 2.6 sec
Substitute with the givens in the equation to get the distance as follows:
s = ut + 0.5a * t^2
s = (0)(2.6) + 0.5(9.8)(2.6)^2
s = 33.124 meters
Hope this helps :)
Answer:
![v_{PB} = 130\ km/h](https://tex.z-dn.net/?f=v_%7BPB%7D%20%3D%20130%5C%20km%2Fh)
Explanation:
Since, Alex is at rest. Therefore, the speed measured by him will be the absolute speed of car P. Therefore, taking easterly direction as positive:
And the absolute velocity of Barbara's Car is given as:![Absolute\ Velocity\ of\ Barbara's\ Car = v_{B} = 52\ km/h](https://tex.z-dn.net/?f=Absolute%5C%20Velocity%5C%20of%5C%20Barbara%27s%5C%20Car%20%3D%20v_%7BB%7D%20%3D%2052%5C%20km%2Fh)
Now, for the velocity of Car p with respect to the velocity of Barbara's Car can be given s follows:
![Velocity\ of\ Car\ P\ measured\ by\ Barbara = v_{PB} = v_{B}-v_{P}\\\\v_{PB} = 52\ km/h-(-78\ km/h)](https://tex.z-dn.net/?f=Velocity%5C%20of%5C%20Car%5C%20P%5C%20measured%5C%20by%5C%20Barbara%20%3D%20v_%7BPB%7D%20%3D%20v_%7BB%7D-v_%7BP%7D%5C%5C%5C%5Cv_%7BPB%7D%20%3D%2052%5C%20km%2Fh-%28-78%5C%20km%2Fh%29)
![v_{PB} = 130\ km/h](https://tex.z-dn.net/?f=v_%7BPB%7D%20%3D%20130%5C%20km%2Fh)