Explanation:
Radius of a charged particle is given by
r=mv / Bq
= k/ q
where k = m v / B is a constant.
i.e. more is the magnitude of charge, less is the radius.
(inversely proportional)
From the diagram r_3 > r_2 > r_1 (more the curvature, less is the radius)
( although drawing is not given i am assuming the above order, however, one can change the order as per the diagram. The concept used remains the same)
therefore, q_1 > q_2 > q_3
.
Answer:
Speed of 1.83 m/s and 6.83 m/s
Explanation:
From the principle of conservation of momentum
where m is the mass, 
is the initial speed before impact, 
and 
are velocity of the impacting object after collision and velocity after impact of the originally constant object
Therefore 
After collision, kinetic energy doubles hence
Substituting 5 m/s for then
Also, it’s known that hence 
Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then 
Substituting, 
Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s
Answer: They are put in front for defense so so they can block the opponents from getting the ball
Explanation:
Answer:
Intensity of beam 18 feet below the surface is about 0.02%
Explanation:
Using Lambert's law
Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium
then dI / I = kdt
taking log,
ln(I) = kt + ln C
I = Ce^kt
t=0=>I=I(0)=>C=I(0)
I = I(0)e^kt
t=3 & I=0.25I(0)=>0.25=e^3k
k = ln(0.25)/3
k = -1.386/3
k = -0.4621
I = I(0)e^(-0.4621t)
I(18) = I(0)e^(-0.4621*18)
I(18) = 0.00024413I(0)
Intensity of beam 18 feet below the surface is about 0.2%
Answer:
v_f = 17.4 m / s
Explanation:
For this exercise we can use conservation of energy
starting point. On the hill when running out of gas
Em₀ = K + U = ½ m v₀² + m g y₁
final point. Arriving at the gas station
Em_f = K + U = ½ m v_f ² + m g y₂
energy is conserved
Em₀ = Em_f
½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂
v_f ² = v₀² + 2g (y₁ -y₂)
we calculate
v_f ² = 20² + 2 9.8 (10 -15)
v_f = √302
v_f = 17.4 m / s
