Answer:
Therefore % increase in velocity is 18.23 %
Explanation:
we use the equality of mass flow rate and the areas
The percentage increase in velocity is
Δ v% = 
100%
= 
.100%
= 
. 100%
= Therefore % increase in velocity is 18.23 %
RT = R1 R2/ R1 + R2
R1 = R2 = 2k ohm
RT = R/2 = 2k/2 = 1k ohm
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
Answer:
Option 3
Explanation:
O Option C is NEGATIVELY CHARGED, meaning it has GAINED ELECTRONS resulting in a GREATER number of ELECTRONS than PROTONS.
