Answer:
b) bend away from the normal
Explanation:
According to snell's law , if i be the angle of incidence and r be the angle of refraction
sin i / sin r = 1.33 / 2.42
sin i / sin r = .55
Hence sin r > sin i
r > i
In other words angle of refraction will be more than angle of incidence . So, the ray will bend away from the normal .
Answer:
The answer is 1.87nm/s.
Explanation:
The
water loss must be replaced by
of sap. 110g of sap corresponds to a volume of
![110g \div \dfrac{1040*10^3g}{1*10^6cm^3} = 106cm^3](https://tex.z-dn.net/?f=110g%20%5Cdiv%20%5Cdfrac%7B1040%2A10%5E3g%7D%7B1%2A10%5E6cm%5E3%7D%20%20%3D%20106cm%5E3)
thus rate of sap replacement is
![106cm^3/hr = 106*10^{-6}m^3/3600s = 2.94*10^{-8}m^3/s](https://tex.z-dn.net/?f=106cm%5E3%2Fhr%20%3D%20106%2A10%5E%7B-6%7Dm%5E3%2F3600s%20%20%3D%202.94%2A10%5E%7B-8%7Dm%5E3%2Fs)
The volume of sap in the vessel of length
is
,
where
is the cross sectional area of the vessel.
For 2000 such vessels, the volume is
![V = 2000Ax](https://tex.z-dn.net/?f=V%20%3D%202000Ax)
taking the derivative of both sides we get:
![\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%20%3D%202000A%20%5Cdfrac%7Bdx%7D%7Bdt%7D)
on the left-hand-side
is the velocity
of the sap, and on right-hand-side
; therefore,
![2.94*10^{-8}m^3/s=2000Av](https://tex.z-dn.net/?f=2.94%2A10%5E%7B-8%7Dm%5E3%2Fs%3D2000Av)
and since the cross-sectional area is
;
therefore,
![2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v](https://tex.z-dn.net/?f=2.94%2A10%5E%7B-8%7Dm%5E3%2Fs%20%3D2000%287.85%2A10%5E%7B-3%7Dm%5E2%29v)
solving for
we get:
![v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}](https://tex.z-dn.net/?f=v%20%3D%20%5Cdfrac%7B2.94%2A10%5E%7B-8%7Dm%5E3%2Fs%7D%7B2000%287.85%2A10%5E%7B-3%7Dm%5E2%29%7D)
![\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}](https://tex.z-dn.net/?f=%5Cboxed%7Bv%20%3D1.875%2A10%5E%7B-9%7Dm%2Fs%20%3D%201.875nm%2Fs%7D)
which is the upward speed of the sap in each vessel.
Answer:
B) t = 1.83 [s]
A) y = 16.51 [m]
Explanation:
To solve this problem we must use the following equation of kinematics.
![v_{f} =v_{o} -g*t](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bo%7D%20-g%2At)
where:
Vf = final velocity = 0
Vo = initial velocity = 18 [m/s]
g = gravity acceleration = 9.81 [m/s²]
t = time [s]
Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.
A) The maximum height is reached when the final velocity of the ball is zero.
0 = 18 - (9.81*t)
9.81*t = 18
t = 18/9.81
t = 1.83 [s], we found the answer for B.
Now using the following equation.
![y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\](https://tex.z-dn.net/?f=y%20%3D%20y_%7Bo%7D%20%2B%20v_%7Bo%7D%2At%20-%200.5%2Ag%2At%5E%7B2%7D%5C%5C)
where:
y = elevation [m]
Yo = initial elevation = 0
y = 18*(1.83) - 0.5*9.81*(1.83)²
y = 16.51 [m]
Answer:
m = B²qR² / 2 V
Explanation:
If v be the velocity after acceleration under potential difference of V
kinetic energy = loss of electric potential energy
1/2 m v² = Vq ,
v² = 2 Vq / m ----------------------- ( 1 )
In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force
magnetic force = centripetal force
Bqv = mv² / R
v = BqR / m
v² = B²q²R² / m² ------------------------- (2)
from (1) and (2)
B²q²R² / m² = 2 Vq / m
m = B²q²R² / 2 Vq
m = B²qR² / 2 V