Answer:
The answer to the question is
Momentum
Explanation:
Momentum in physics refer to the attribute that a body has by virtue of its mass and velocity. It is found by multiplying the mass of the moving object and the velocity, hence
Momentum = Mass × Velocity = m·v
Newton first law of motion states that the force acting on an object is proportional to the rate of change of momentum produced
Therefore when we find the momentum of the two cars, the one that has the greater momentum will require the most force to stop it.
Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team that is on the move has the momentum. If an object is in motion (on the move) then it has momentum
Momentum
The amount of fluid that moves past a point in area A per unit of time is known as the flow rate.
<h3>How do you find average velocity from flow rate?</h3>
- The amount of fluid that moves past a point in area A per unit of time is known as the flow rate. Here, a uniform pipe carrying the shaded fluid cylinder passes point P in time t. The cylinder's capacity is Ad, its average velocity is v=d/t, and its flow rate is Q=Ad/t=Av.
- The average fluid velocity for laminar flow through a pipe is equal to half of the fluid's greatest velocity at the pipe's center. The Hagen-Poiseuille equation is shown above. Since there is no acceleration in a steady and uniform flow, there is no force acting in the direction of the flow.
To learn more about fluid velocity refer,
brainly.com/question/13344039
#SPJ4
3.) Air
4.) Stay same size but different internal pressure
5.) Stars
Answer:
order d> a = e> c> b = f
Explanation:
Pascal's law states that a change in pressure is transmitted by a liquid, all points are transmitted regardless of the form
P₁ = P₂
Using the definition of pressure
F₁ / A₁ = F₂ / A₂
F₂ = A₂ /A₁ F₁
Now we can examine the results
a) F1 = 4.0 N A1 = 0.9 m2 A2 = 1.8 m2
F₂ = 1.8 / 0.9 4
F₂a = 8 N
b) F1 = 2.0 N A1 = 0.9 m2 A2 = 0.45 m2
F₂b = 0.45 / 0.9 2
F₂b = 1 N
c) F1 2.0 N A1 = 1.8 m2 A2 = 3.6 m2
F₂c = 3.6 / 1.8 2
F₂c = 4 N
d) F1 = 4.0N A1 = 0.45 m2 A2 = 1.8 m2
F₂d = 1.8 / 0.45 4.0
F₂d = 16 m2
e) F1 = 4.0 N A1 = 0.45 m2 A2 = 0.9 m2
F₂e = 0.9 / 0.45 4
F₂e = 8 N
f) F1 = 2.0N A1 = 1.8 m2 A2 = 0.9 m2
F₂f = 0.9 / 1.8 2.0
F₂f = 1 N
Let's classify the structure from highest to lowest
F₂d> F₂a = F₂e> F₂c> F₂b = F₂f
I mean the combinations are
d> a = e> c> b = f
Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
= t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s