Answer:
initial velocity is v = 4.95 m / s
Explanation:
To solve this exercise we use the projectile launch ratios, when the block leaves the hill its speed is horizontal, let's find the time it takes to fall to the other point.
Initial vertical velocity is zero
y = y₀ + v_{oy} t - ½ g t²
y-y₀ = 0 -1/2 g t²
t = 
calculate
t = 
t = 2.02 s
with this time we can substitute in the horizontal displacement equation
x = v₀ₓ t
v₀ₓ = x / t
suppose that the distance between the two points is x = 10 m
v₀ₓ = 10 / 2.02
v₀ₓ = 4.95 m / s
initial velocity is v = 4.95 m / s
Fly in a straight line unless an outside force changes its course because i tried it once in a baseball game that my mommy rekt me in.
Answer:
Keq = 2k₃
Explanation:
We can solve this exercise using Newton's second one
F = m a
Where F is the eleatic force of the spring F = - k x
Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement
F + F = m a
k₃ x + k₃ x = m a
a = 2k₃ x / m
To find the effective force constant, suppose we change this spring to what creates the cuddly displacement
Keq = 2k₃
The acceleration of the object which moves from an initial step to a full halt given the distance traveled can be calculated through the equation,
d = v² / 2a
where d is distance, v is the velocity, and a is acceleration
Substituting the known values,
180 = (22.2 m/s)² / 2(a)
The value of a is equal to 1.369 m/s²
The force needed for the object to be stopped is equal to the product of the mass and the acceleration.
F = (1300 kg)(1.369 m/s²)
F = 1779.7 N
