Answer:
the velocity is 10 m/s
Explanation:
Using the expression for kinetic energy we have:
![Ek=\frac{1}{2} *m*v^{2} \\\\Ek=100J\\m=2kg\\v=\sqrt{(2*100/2)}\\ v=10[m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5C%5C%5CEk%3D100J%5C%5Cm%3D2kg%5C%5Cv%3D%5Csqrt%7B%282%2A100%2F2%29%7D%5C%5C%20v%3D10%5Bm%2Fs%5D)
Answer:
We kindly invite you to read carefully the explanation and check the image attached below.
Explanation:
According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:
(1)
Where:
- Initial velocity, measured in meters per second.
- Final velocity, measured in meters per second.
- Acceleration, measured in meters per square second.
- Initial time, measured in seconds.
- Final time, measured in seconds.
Now we obtain the kinematic equations for thrust and free fall stages:
Thrust (
,
,
,
)
(2)
Free fall (
,
,
,
)
(3)
Now we created the graph speed-time, which can be seen below.
Answer:
D
friction acts in the opposite direction of motion but does not affect the motion of the object
Dipper effect of an oncoming train get louder as it approaches and sound diminishes as it goes away sound traveling
<u>Answer:</u>
<em>Thunderbird is 995.157 meters behind the Mercedes</em>
<u>Explanation:</u>
It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.
Here final velocity ![v=0 \ m/s](https://tex.z-dn.net/?f=v%3D0%20%5C%20m%2Fs)
initial velocity
distance
Distance covered in the slowing down phase = ![250 m](https://tex.z-dn.net/?f=250%20m)
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
![a= \frac {(v^2-u^2)}{2s}](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%20%7B%28v%5E2-u%5E2%29%7D%7B2s%7D)
![a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%20%7B%280%5E2-71%5E2%29%7D%7B%282%20%5Ctimes%20250%29%7D%3D-10.082%20%5C%20m%2Fs%5E2)
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
![t= \frac {(v-u)}{a}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%20%7B%28v-u%29%7D%7Ba%7D)
![= \frac {(0-71)}{(-10.082)}=7.042 s](https://tex.z-dn.net/?f=%3D%20%5Cfrac%20%7B%280-71%29%7D%7B%28-10.082%29%7D%3D7.042%20s)
![t_1=7.042 s](https://tex.z-dn.net/?f=t_1%3D7.042%20s)
The car is in the pit stop for 5s ![t_2=5 s](https://tex.z-dn.net/?f=t_2%3D5%20s)
After restart it accelerates for 350 m to reach the earlier velocity 71 m/s
![a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%20%7B%28v%5E2-u%5E2%29%7D%7B%282%5Ctimes%20s%29%7D%20%3D%20%5Cfrac%7B%2871%5E2-0%5E2%29%7D%7B%282%20%5Ctimes%20370%29%7D%20%3D6.81%20%5C%20m%2Fs%5E2)
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
![t= \frac{(v-u)}{a}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B%28v-u%29%7D%7Ba%7D)
![t= \frac{(71-0)}{6.81}= 10.425 s](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B%2871-0%29%7D%7B6.81%7D%3D%2010.425%20s)
![t_3=10.425 s](https://tex.z-dn.net/?f=t_3%3D10.425%20s)
total time= ![t_1 +t_2+t_3=7.042+5+10.425=22.467 s](https://tex.z-dn.net/?f=t_1%20%2Bt_2%2Bt_3%3D7.042%2B5%2B10.425%3D22.467%20s)
Distance covered by the Mercedes Benz during this time is given by ![s=vt=71 \times 22.467= 1595.157 m](https://tex.z-dn.net/?f=s%3Dvt%3D71%20%5Ctimes%2022.467%3D%201595.157%20m)
Distance covered by the Thunderbird during this time=![250+350=600 m](https://tex.z-dn.net/?f=250%2B350%3D600%20m)
Difference between distance covered by the Mercedes and Thunderbird
= ![1595.157-600=995.157 m](https://tex.z-dn.net/?f=1595.157-600%3D995.157%20m)
Thus the Mercedes is 995.157 m ahead of the Thunderbird.