Acceleration = (change in speed) / (time for the change)
Change in speed = (speed at the end) minus (speed at the beginning.
The cart's acceleration is
(0 - 2 m/s) / (0.3 sec)
= ( -2 / 0.3 ) (m/s²) = -(6 and 2/3) m/s² .
Newton's second law of motion says
Force = (mass) x (acceleration) .
For this cart: Force = (1.5 kg) x ( - 6-2/3 m/s²)
= ( - 1.5 x 20/3 ) (kg-m/s²)
<span> = </span>- 10 newtons .
<span>The force is negative because it acts opposite to the direction </span>
<span>in which the cart is moving, it causes a negative acceleration, </span>
<span>and it eventually stops the cart.</span>
The angle is found by taking the arctangent of the vertical angle divided by the horizontal angle
θ = arctan(18/25) = 35.75°
36° is the closest answer.
Answer:
Option D is the correct answer.
Explanation:
We equation for elongation
Here we need to find weight required,
We need to stretch a steel road by 2 mm, that is ΔL = 2mm = 0.002 m

E = 200GPa = 2 x 10¹¹ N/m²
L=2m
Substituting

Option D is the correct answer.
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