Answer:
speed = 3.95 m/s
Explanation:
area = π x radius^2
area = π x (2.67 x 10^-3)^2
volume flow rate = area x speed
volume / time = area x speed
density = mass / volume
volume = mass / density
<u>mass / (density x time) = area *speed</u>
mass flow rate = mass / time
<u>mass flow rate / density = area x speed</u>
6.55 x 10^-2 / 740 = pi * (2.67 x 10^-3)^2 * speed
speed =8.8514 x 10-5 /2.2396 x 10-5 m/s
speed = 3.95 m/s
Answer:
d_2 = 4d_1
Explanation:
The range or horizontal distance covered by a projectile projected with a velocity U at an angel of θ to the horizontal is given by
R = U²sin2θ/g
Let the range or horizontal distance of ball 1 with initial velocity U projected at an angle θ = 55° be
d_1 = U²sin2θ/g
Let the range or horizontal distance of ball 2 with initial velocity V = 2U projected at an angle θ = 55° be
d_2 = V²sin2θ/g
= (2U)²sin2θ/g
= 4U²sin2θ/g
= 4d_1 (since d_1 = U²sin2θ/g)
So, the ball 2 lands a distance d_2 = 4d_1 from the initial point.
Answer:
peat them with my dislocated legs
The most important measure is awhips
vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6
Therefore, v= 3.5 m/s.
