Answer:
V0=27.4 m/s; t=0.8 s
Explanation:
Final position y=37.0 m, time = 2.3 s; Initial position is set to be zero. We calculate the initial speed with the kinematics equation:
We solve for initial speed

Now, using the same expression we estimated time to first reach 18.5 m :
Second order equation with solutions
t1=0.8 s and t2=4.8 s
The first time corresponds to the first reach.
Answer:
The speed of light will be c=3x10^8m/s
Explanation:
This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c
<h2>Answer:</h2>
The refractive index is 1.66
<h2>Explanation:</h2>
The speed of light in a transparent medium is 0.6 times that of its speed in vacuum
.
Refractive index of medium = speed of light in vacuum / speed of light in medium
So
RI = 1/0.6 = 5/3 or 1.66
I assume here that the engine operates following a Carnot cycle, which achieves the maximum possible efficiency.
Under this assumption, the efficiency of the engine (so, the efficiency of the Carnot cycle) is given by

where

is the cold temperature

is the hot temperature
For the engine in our problem, the cold temperature is 313 K while the hot temperature is 425 K, so the effiency of the engine is