Answer:
spring deflection is x = (v2 / R + g) m / 4
Explanation:
We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration
Let's write the equations on the Y axis of this description
Fe - W = m 
Where Fe is elastic force, W the weight and the centripetal acceleration. The elastic force equation is
Fe = - k x
4 (k x) - mg = m v² / R
The four is because there are four springs, R is theradio of dip
We can calculate the deflection (x) of the springs
x = (m v2 / R + mg) / 4
x = (v2 / R + g) m / 4
Answer:
well, the hill isn't constantly going down hill, there's an ending point or goes back up hill making a v/u shape or there's nothing helping the wagon being pushed or pull currently
Explanation:
Answer:
Zero
Explanation:
W = F × s
F = 10 N,
t = 3min = 180sec
s = 0( no change in postion)
W = 10 ×0
W = 0
Answer:
10kg
Explanation:
Let PE=potential energy
PE=196J
g(gravitational force)=9.8m/s^2
h(change in height)=2m
m=?
PE=m*g*(change in h)
196=m*9.8*2
m=10kg
Answer:
The minimum possible coefficient of static friction between the tires and the ground is 0.64.
Explanation:
if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :
Fc = f
m×(v^2)/(R) = μ×m×g
(v^2)/(R) = g×μ
μ = (v^2)/(R×g)
= ((25)^2)/((100)×(9.8))
= 0.64
Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.
