That's false. No medium = no sound.
The answer is 35 degrees Celsius. Hope I helped :) Please vote brainliest.
Answer:
The K.E is maximum when the child is at the vertical position and the P.E is maximum at the extreme deviated position from the vertical.
Explanation:
- A child is swinging on swing up and down has both kinetic and potential energy.
- The total mechanical energy of the system is conserved throughout the system. At any instant the total mechanical energy is given by,
E = K.E + P.E
- The K.E is maximum when the child is at the vertical position.
- The P.E is maximum at the extreme deviated position from the vertical.
- And when K.E is maximum P.E becomes minimum and vice versa as per the law of conservation of energy.
Answer: It represents the whole distance traveled. Hope this helps!
Explanation:
Incomplete question as there is so much information is missing.The complete question is here
A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?
Answer:
Distance traveled=240 m
Explanation:
Given data
Initial velocity of car v₀=0 m/s
Final velocity of car vf=24 m/s
Distance traveled by car S=120 m
To find
Distance does the traffic travel
Solution
To find the distance first we need to find time, for time first we need acceleration
So
![(V_{f})^{2}=(V_{o})^{2}+2aS\\ So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}](https://tex.z-dn.net/?f=%28V_%7Bf%7D%29%5E%7B2%7D%3D%28V_%7Bo%7D%29%5E%7B2%7D%2B2aS%5C%5C%20%20So%5C%5Ca%3D%5Cfrac%7B%28V_%7Bf%7D%29%5E%7B2%7D-%28V_%7Bo%7D%29%5E%7B2%7D%20%7D%7B2S%7D%5C%5C%20a%3D%5Cfrac%7B%2824m%2Fs%29%5E%7B2%7D-%280m%2Fs%29%5E%7B2%7D%20%7D%7B2%28120%29%7D%5C%5Ca%3D2.4%20m%2Fs%5E%7B2%7D)
As we find acceleration.Now we need to find time
So
![V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s](https://tex.z-dn.net/?f=V_%7Bf%7D%3DV_%7Bi%7D%2Bat%5C%5Ct%3D%5Cfrac%7BV_%7Bf%7D-V_%7Bi%7D%7D%7Ba%7D%5C%5Ct%3D%5Cfrac%7B%2824m%2Fs%29-%280m%2Fs%29%7D%7B%282.4m%2Fs%5E%7B2%7D%20%29%7D%5C%5Ct%3D10s)
Now for distance
So
![Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m](https://tex.z-dn.net/?f=Distance%3Dvelocity%2Atime%5C%5CDistance%3D%2824m%2Fs%29%2A%2810s%29%5C%5CDistance%3D240m)