Answer:
<em>The speed of the projectile when it impacts the ground is 1000 m/s</em>
Explanation:
<u>Vertical Launch</u>
When an object is launched vertically and upwards it starts to move at an initial speed vo, then the acceleration of gravity makes that speed to reduce until it reaches 0. The object has reached its maximum height. Then, it starts to move downwards in free fall, with initial speed zero and gradually increasing it until it reaches the ground level. We will demonstrate that the speed it has when impacts the ground is the same (and opposite) as the initial speed vo.
The speed when the object is moving upwards is given by
The time it takes to reach the maximum height is when vf=0, i.e.
solving for t
The maximum height reached is
Then, the object starts to fall. The object's height is given by
where t' is the time the object has traveled downwards. The height will be 0 again when
Solving for t'
We can see the time it takes to reach the maximum height is the same it takes to return to ground level. Of course, the speed when it happens is
Thus, the speed of the projectile when it impacts the ground is 1000 m/s
When storm clouds produce lightening and thunder, then, potential energy changes into kinectic energy.
The clouds become negatively charged,and the ground is positively charged during a storm. So there creates a potential difference between the charged clouds and the ground. And when this becomes huge so much that it over comes the electrical insulation cover of air, the electrons jump to the ground.i.e. the potential energy changes into kinetic energy.
Answer:
a) k= 3232.30 N / m, b) f = 4,410 Hz
Explanation:
In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.
The expression for the angular velocity is
w = √k/m
the angular velocity is related to the period
w = 2π / T
we substitute
T = 2
√m/ k
a) empty car
k = 4π² m / T²
k = 4 π² 1310/2 2
k = 12929.18 N / m
This is the equivalent constant of the short springs
F1 + F2 + F3 + F4 = k_eq x
k x + kx + kx + kx = k_eq x
k_eq = 4 k
k = k_eq / 4
k = 12 929.18 / 4
k= 3232.30 N / m
b) the frequency of oscillation when carrying four passengers.
In this case the plus is the mass of the vehicle plus the masses of the passengers
m_total = 1360 + 4 70
m_total = 1640 kg
angular velocity and frequency are related
w = 2pi f
we substitute
2 pi f = Ra K / m
in this case the spring constant changes us
k_eq = 12929.18 N / m
f = 1 / 2π √ 12929.18 / 1640
f = π / 2 2.80778
f = 4,410 Hz
