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:
Given
Lowest four resonance frequencies are given with magnitude
50,100,150 and 200 Hz
The frequency of vibrating string is given by

where n=1,2,3 or ...n
L=Length of string
T=Tension
Mass per unit length
When string is clamped at mid-point
Effecting length becomes 
Thus new Frequency becomes

i.e. New frequency is double of old
so new lowest four resonant frequencies are 100,200,300 and 400 Hz
Answer:
No, there won't be a collision.
Explanation:
We will use the constant acceleration formulas to calculate,
v = u + a*t
0 = 25 + (-0.1)*t
t = 250 seconds (the time taken for the passenger train to stop)
v^2 = u^2 + 2*a*s
0 = (25)^2 + 2*(-0.1)*s
s = 3125 m (distance traveled by passenger train to stop)
If the distance traveled by freight train in 250 seconds is less than (3125-200=2925 m) than the collision will occur
Speed*time = distance
Distance = (15)*(250)
Distance = 3750 m
As the distance is way more, there won’t be a collision
Physical and chemical changes are similar because matter experiences a change in state while undergoing either change.