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:
Answer:
I believe this is German for something day how are you????? I was talking about the holocaust again in school and decided to start learning German so if im wrong then.. wowww!
Explanation:
Newton’s Second Law concerns the generation of force based on an object’s mass and acceleration, as described by the equation F=ma.
Hope this helps!
Answer: Part(a)=0.041 secs, Part(b)=0.041 secs
Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction
now we know that
a=-9.81 m/s^2 ( negative because it is pulling the player downwards)
we also know that
s=76 cm= 0.76 m ( maximum s)
using kinetic equation
where v is final velocity which is zero at max height and u is it initial
hence
now we can find time in the 15 cm ascent
using quadratic formula
t=0.0409 sec
the answer for the part b will be the same
To find the answer for the part b we can find the velocity at 15 cm height similarly using
where s=0.76-0.15
as the player has traveled the above distance to reach 15cm to the bottom
when the player reaches the bottom it has the same velocity with which it started which is 3.861
hence the time required to reach the bottom 15cm is
t=0.0409
Answer:
(B) 0.5 g
Explanation:
Newton's second law says ∑ F i = m a .
the rate of change in momentum of a body is proportional to the force applied on the body.
f∝ma
f=kma
were k is constant and equal to 1
The centripetal acceleration is an acceleration.
the tension on the swing and object weight goes to the left hand side while the centripetal acceleration goes to the right handside
At the bottom of the swing, ΣF = FT – mg = mac;
notice that the tension in the swing is 1.5 times the weight of the object
we can write
1.5mg – mg = mac,
0.5mg = mac
0.5 g=ac
