Answer:
The water will flow at a speed of 3,884 m/s
Explanation:
Torricelli's equation
v =
*v = liquid velocity at the exit of the hole
g = gravity acceleration
h = distance from the surface of the liquid to the center of the hole.
v = = 3,884 m/s
Answer:
Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m
Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m
Explanation:
Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.
H = (v₀² Sin²θ)/2g
v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s
θ = 75°, g = 9.8 m/s²
H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m
Range of projectile
R = v₀² (sin2θ)/g
R = 250² (sin2×75)/9.8
R = 250² (sin 150)/9.8 = 3188.8 m
Height of mountain = 1.80 × 10³ = 1800 m
Maximum height of projectile = 2975.2 m
Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m
Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m
Range of projectile = 3188.8 m
Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m
Explanation:
Given that,
Angular velocity = 0.240 rev/s
Angular acceleration = 0.917 rev/s²
Diameter = 0.720 m
(a). We need to calculate the angular velocity after time 0.203 s
Using equation of angular motion
Put the value in the equation
The angular velocity is 0.426 rev/s.
(b). We need to calculate the tangential speed of the blade
Using formula of tangential speed
Put the value into the formula
The tangential speed of the blade is 0.963 m/s.
(c). We need to calculate the magnitude at of the tangential acceleration
Using formula of tangential acceleration
Put the value into the formula
The tangential acceleration is 2.074 m/s².
Hence, This is required solution.