Answer:
kick 1 has travelled 15 + 15 = 30 yards before hitting the ground
so kick 2 travels 25 + 25 = 50 yards before hitting the ground
first kick reached 8 yards and 2nd kick reached 20 yards
Explanation:
1st kick travelled 15 yards to reach maximum height of 8 yards
so, it has travelled 15 + 15 = 30 yards before hitting the ground
2nd kick is given by the equation
y (x) = -0.032x(x - 50)

we know that maximum height occurs is given as


and maximum height is

y = 20
so kick 2 travels 25 + 25 = 50 yards before hitting the ground
first kick reached 8 yards and 2nd kick reached 20 yards
Answer:Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance between two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is 340 m/s
Explanation:
Answer:
a) p = m1 v1 + m2 v2
, b) dp / dt = m1 a1 + m2 a2
, c) It is equivalent to force
dp / dt = 0
Explanation:
In this problem we have two blocks and the system is formed by the two bodies.
Part A. Initially they ask us to find the moment of the whole system
p = m1 v1 + m2 v2
Part B.
Find the derivative
dp / dt = m1 dv1dt + m2 dv2 / dt
dp / dt = m1 a1 + m2 a2
Part C.
Let's analyze the dimensions
m a = [kg] [m / s2] = [N]
It is equivalent to force
Part d
Acceleration is due to a net force applied
Part e
The acceleration of block 1 is due to the force exerted by block 2 during the moment change
Part f
Force of block 1 on block 2
True f12 = m1a1 f21 = m2a2
Part g
By the law of action and reaction are equal magnitude F12 = f21
Part H
dp / dt = 0
Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed
The data for the first part of the experiment support the first hypothesis. As the force applied to the cart increased, the acceleration of the cart increased. Since the increase in the applied force caused the increase in the cart's acceleration, force and acceleration are directly proportional to each other, which is in accordance with Newton's second law.
Answer:
a) v = 1.19 m / s
, b) P₁ = 0.922 10⁵ Pa
Explanation:
1) Let's use the fluid continuity equation
Q = A v
The area of a circle is
A = π r2 = π d²/4
v = Q / A = Q 4 / pi d²
v = 0.006 4/π 0.08²
v = 1.19 m / s
2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point
P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂
The exercise tell us
P₂ = 1.0013 105 Pa
v₁ = 0
y₁ = 1 m
y₂=0
Rho (water) = 1000 kg / m³
P₁ + rho y₁ = P₂ + ½ rho v₂²
P₁ = P₂ + ½ rho v₂² - rho g y₁
P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1
P₁ = 1.013 10⁵ +708.5 - 9800
P₁ = 92208.5Pa
P₁ = 0.922 10⁵ Pa