Answer: ![1.23\ m/s^2](https://tex.z-dn.net/?f=1.23%5C%20m%2Fs%5E2)
Explanation:
Given
At an elevation of
, spacecraft is dropping vertically at a speed of ![u=293\ m/s](https://tex.z-dn.net/?f=u%3D293%5C%20m%2Fs)
Final velocity of the spacecraft is ![v=0](https://tex.z-dn.net/?f=v%3D0)
using equation of motion i.e. ![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
Insert the values
![\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2](https://tex.z-dn.net/?f=%5CRightarrow%200-%28293%29%5E2%3D2%5Ctimes%20a%5Ctimes%20%2834.7%5Ctimes%2010%5E3%29%5C%5C%5C%5C%5CRightarrow%20a%3D-%5Cdfrac%7B293%5E2%7D%7B2%5Ctimes%2034.7%5Ctimes%2010%5E3%7D%5C%5C%5C%5C%5CRightarrow%20a%3D-1.23%5C%20m%2Fs%5E2)
Therefore, magnitude of acceleration is
.
As per energy conservation we know that
Energy enter into the bulb = Light energy + Thermal energy
so now we have
energy enter into the bulb = 100 J
Light energy = 5 J
now from above equation we have
![100 = 5 + heat](https://tex.z-dn.net/?f=100%20%3D%205%20%2B%20heat)
![Heat = (100 - 5) J](https://tex.z-dn.net/?f=Heat%20%3D%20%28100%20-%205%29%20J)
![Heat = 95 J](https://tex.z-dn.net/?f=Heat%20%3D%2095%20J)
Answer:
Kinetic energy is 242.7 Joule
Explanation:
m = 47.4 kg
v = 3.2m/s²
Now,
K.E. = 1/2 × m × v²
or, K.E. = 1/2 × 47.4 × 3.2²
or, K.E. = 242.7 Joule
Answer:
(a). 19.0 m/s
(b). 9.5m/s
Explanation:
Assuming speed of sound is 343m/s.
(a).
As the train approaches, from the the Doppler equation we have
![f_{obs} = \dfrac{vf_{source}}{v-v_{source}}](https://tex.z-dn.net/?f=f_%7Bobs%7D%20%3D%20%5Cdfrac%7Bvf_%7Bsource%7D%7D%7Bv-v_%7Bsource%7D%7D)
![350Hz = \dfrac{343f_{source}}{343-v_{source}}](https://tex.z-dn.net/?f=350Hz%20%3D%20%5Cdfrac%7B343f_%7Bsource%7D%7D%7B343-v_%7Bsource%7D%7D)
solving for
we get:
.
And as the train cuts its speed in half, the equation gives
![340Hz = \dfrac{343f_{source}}{343-\dfrac{v_{source}}{2} }](https://tex.z-dn.net/?f=340Hz%20%3D%20%5Cdfrac%7B343f_%7Bsource%7D%7D%7B343-%5Cdfrac%7Bv_%7Bsource%7D%7D%7B2%7D%20%7D)
substituting the value of
we get:
![340Hz = \dfrac{343\dfrac{350(343-v_{source})}{343}}{343-\dfrac{v_{source}}{2} }](https://tex.z-dn.net/?f=340Hz%20%3D%20%5Cdfrac%7B343%5Cdfrac%7B350%28343-v_%7Bsource%7D%29%7D%7B343%7D%7D%7B343-%5Cdfrac%7Bv_%7Bsource%7D%7D%7B2%7D%20%7D)
![340Hz = \dfrac{350(343-v_{source})}{343-\dfrac{v_{source}}{2} }](https://tex.z-dn.net/?f=340Hz%20%3D%20%5Cdfrac%7B350%28343-v_%7Bsource%7D%29%7D%7B343-%5Cdfrac%7Bv_%7Bsource%7D%7D%7B2%7D%20%7D)
![340Hz = \dfrac{700(343-v_{source})}{686-v_{source}}](https://tex.z-dn.net/?f=340Hz%20%3D%20%5Cdfrac%7B700%28343-v_%7Bsource%7D%29%7D%7B686-v_%7Bsource%7D%7D)
![340 (686-v_{source}) = 700(343-v_{source})](https://tex.z-dn.net/?f=340%20%28686-v_%7Bsource%7D%29%20%3D%20700%28343-v_%7Bsource%7D%29)
![\boxed{v_{source} = 19.0m/s}](https://tex.z-dn.net/?f=%5Cboxed%7Bv_%7Bsource%7D%20%3D%2019.0m%2Fs%7D)
or 68.6 km per hour, which is the speed of the train before slowing down.
(b).
The speed of the train after slowing down is half the previous speed; therefore,
![v_{after} = \dfrac{v_{source}}{2}](https://tex.z-dn.net/?f=v_%7Bafter%7D%20%3D%20%5Cdfrac%7Bv_%7Bsource%7D%7D%7B2%7D)
![\boxed{v_{after} =9.5m/s}](https://tex.z-dn.net/?f=%5Cboxed%7Bv_%7Bafter%7D%20%3D9.5m%2Fs%7D)
or 34.3 km per hour.