W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s
Newton's laws A force cannot act alone is the THIRD LAW!
Answer:
the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south
Explanation:
given information:
Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus
A= 27
Ax = 27 sin 60 = - 23.4
Ay = 27 cos 60 = 13.5
Jane walks 16.0 m in a direction 30.0 ∘ south of west, so
B = 16
Bx = 16 cos 30 = -13.9
By = 16 sin 30 = -8
the direction that should be walked by Ricardo to go directly to Jane
R = √A²+B² - (2ABcos60)
= √27²+16² - (2(27)(16)(cos 60))
= 23.52 m
now we can use the sines law to find the angle
tan θ = 
= By - Ay/Bx -Ax
= (-8 - 13.5)/(-13.9 - (-23.4))
θ = 90 - (-8 - 13.5)/(-13.9 - (-23.4))
= 24° east of south
Answer:
Water flowing rate= (300000kg/s) = (300000l/s)
Explanation:
First with the section of the channel, the depth of the water and the speed of the fluid we can determine the volume of fluid that circulates per second through the channel:
Volume per time= 15m × 8m × (2.5m/s)= 300 m³/s
With this volume of circulating fluid per second elapsed, we multiply it by the density of the water to determine the kilograms or liters of water that circulate through the channel per second elapsed:
Water flowing rate= (300m³/s) × (1000kg/m³)= (300000kg/s) = (300000l/s)
Taking into account that 1kg of water is approximately equal to 1 liter of water.