Cell phones use RF waves from nearby cell towers. RF waves fall in between FM radio waves and microwaves
Using the two kinematic equations that can be used for this problem are:
Vf = Vi + at and d=Vit +(1/2)*at^2
Since Vi (initial velocity) = 0
The equations can further be simplified where a is the acceleration, t is the time, Vf is the final velocity which is 70 miles per hour and d is 6 miles
Vf = at
70 = at
a = 70/t---equation 1
d=(1/2)*a*(t^2)
6 = (1/2)*a*(t^2) ---equation 2
Substituting equation 1 to equation 2.
6= (1/2)*(70/t)*(t^2)
6= 35t
t= 0.17142 hours or 10.28571 mins or 617.14 sec
Answer:
distance between the two second-order minima is 2.8 cm
Explanation:
Given data
distance = 1.60 m
central maximum = 1.40 cm
first-order diffraction minima = 1.40 cm
to find out
distance between the two second-order minima
solution
we know that fringe width = first-order diffraction minima /2
fringe width = 1.40 /2 = 0.7 cm
and
we know fringe width of first order we calculate slit d
β1 = m1λD/d
d = m1λD/β1
and
fringe width of second order
β2 = m2λD/d
β2 = m2λD / ( m1λD/β1 )
β2 = ( m2 / m1 ) β1
we know the two first-order diffraction minima are separated by 1.40 cm
so
y = 2β2 = 2 ( m2 / m1 ) β1
put here value
y = 2 ( 2 / 1 ) 0.7
y = 2.8 cm
so distance between the two second-order minima is 2.8 cm
Answer:
20N
Explanation:
The force (Hooke's Law) is proportional to the deformation. To extend the spring 4 times longer, you will need 4 times the force. In total, 20 N
