Answer:
a) Team A will win.
b) The losing team will accelerate towards the middle line with 0.01 m/
Explanation:
Given that Team-A pulls with a force , 
and Team-B pulls with a force , 
∵ 
The rope will move in the direction of force 
.
∴ Team-A will win.
b) Considering both the teams as one system of total mass , 
Net force on the system , 
= 50-45 = 5N
Applying Newtons first law to the system ,
F = ma , where 'a' is the acceleration of the system.
Since , both the teams are connected by the same rope , their acceleration would be the same.
∴ 5 = 499×a
∴ a = 0.01 m/
Radio waves are the lowest level of waves
In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
(1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit
In our problem,
while the distance between the first and the fifth minima is
(2)
If we use the formula to rewrite
, eq.(2) becomes
Which we can solve to find a, the width of the slit:
It’s more dense than air and less dense than liquid!
F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N
