Answer:
78.4 KN/m
Explanation:
Given
mass of person 'm' =80 kg
car dips about i.e spring stretched 'x'= 1 cm => 0.01m
acceleration due to gravity 'g'= 9.8 m/s^2
as we know that,in order to find approximate spring constant we use Hooke's Law i.e F=kx
where,
F = the force needed
x= distance the spring is stretched or compressed beyond its natural length
k= constant of proportionality called the spring constant.
F=kx ---> (since f=mg)
mg=kx
k=(mg)/x
k=(80 x 9.8)/ 0.01
k=78.4x
k=78.4 KN/m
An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7
In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.
As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.
Using the Newton second law;
N = mg cos θ
Equating both force component together, we have:
From trigonometry rule:
∴
Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7
Learn more about static friction here:
brainly.com/question/24882156?referrer=searchResults
Given that,
Mass of a tribble, m = 2.5 kg
Radius, r = 1.4 m
The force on the tribble from the bucket does not exceed 10 times its weight.
To find,
The maximum tangential speed.
Solution,
The force acting on the tribble is equal to the centripetal force.
F = 10mg
The formula for the centripetal force is given by :
v is maximum tangential speed
So, the maximum tangential speed is 11.7 m/s.