Answer:
The width of the slit is 0.167 mm
Explanation:
Wavelength of light, 
Distance from screen to slit, D = 88.5 cm = 0.885 m
The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m
We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :
where
a = width of the slit
a = 0.000167 m
a = 0.167 mm
So, the width of the slit is 0.167 mm. Hence, this is the required solution.
Answer:
Explanation:
As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from 
to 
is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.
The total distance covered by the train during the entire journey is the area of the speed-time graph.
Area
As velocity is in 
and time is in 
so the unit of area is 
Hence, the total distance is 
.
Answer:
I'm not 100% sure, but I think the answer would be the first one because there's a force pushing the object in every direction, so they would cancel eachother out and make the object stay in the same place.
Explanation:
pls vote brainliest
Answer:
a = 3 m/s^2
Explanation:
Vi = 10 m/s
Vf = 40 m/s
t = 10 s
Plug those values into the following equation:
Vf = Vi + at
40 = 10 + 10a
---> a = 3 m/s^2
