Answer:
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
Explanation:
initial veetical speed V₀y=0
Horizontal speed Vx = Vx₀= 3.80m/s
Vertical drop height= 3.90m
Let Vy = vertical speed when it got to the water downward.
g= 9.81m/s² = acceleration due to gravity
From kinematics equation of motion for vertical drop
Vy²= V₀y² +2 gh
Vy²= 0 + ( 2× 9.8 × 3.90)
Vy= √76.518
Vy=8.747457
Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below
V= √Vy² + Vx²
V=√3.80² + 8.747457²
V=9.537m/s
The angle can also be calculated as
θ=tan⁻¹(Vy/Vx)
tan⁻¹( 8.747457/3.80)
=66.52⁰
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
d=vi*t+(1/2)gt²
d=11 m
g=9.8 m/s²
vi=0 m/s
11 m=0 m/s*t+(1/2)9.8 m/s²t²
11 m=4.9 m/s²t²
t²=11 m / 4.9 m/s²
t=√(11 m / 4.9 m/s²)=1.489... s≈1.5 s
Answer: the time the sone is in flight is 1.5 s
Answer: Your nose inhales O2 and goes through these little nose hairs to keep stuff that doesn't belong in your airway, it goes through your nose into your lungs taking in O2 for your blood. The other substances that you have inhaled get absorbed elsewhere and it is converted into CO2 which is what you exhale.
Explanation:
Answer:
7.00 m
Explanation:
Given:
v₀ = 2.00 m/s
v = 5.00 m/s
a = 1.50 m/s²
Find: Δx
v² = v₀² + 2aΔx
(5.00 m/s)² = (2.00 m/s)² + 2(1.50 m/s²)Δx
Δx = 7.00 m
Answer:
Given:
m=1000kg
u= 16.7m/s
v=0m/s
F=8000N
Required:
s=?
Solution:
F=m × a
8000N=1000kg × a
a=8m/s^2
Since it decelerate a= -8m/s^2
v^2 = u^2 + 2as
s=v^2 - u^2 / 2a
s= 0 - (16.7m/s)^2 / 2 × -8m/s^2
s= -278.89/-16
s= 17.43m
The car travels approximately 17.43m before it stops
Please like and follow me
