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Answer:
0.739
Explanation:
If we treat the four tire as single body then
W ( weight of the tyre ) = mass × acceleration due to gravity (g)
the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r
where v is speed 25.6 m/s and r is the radius of the circle
centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²
net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²
coefficient of static friction between the tires and the road = frictional force / force of normal
frictional force = m × net acceleration / m×g
where force of normal = weight of the body in opposite direction
coefficient of static friction = (7.2567 × m) / (9.81 × m)
coefficient of static friction = 0.739
Answer:
the googles are 4.33mm from the edge
Explanation:
In this question, we are asked to calculate the distance of a set of googles from the edge of a pool.
We proceed as follows;
depth of pool , d = 3.1mm
Now, let i be the angle of incidence
i = arctan(1.8/1.2)
i = 56.31 degree
Using snell's law ,
n1 * sin(i) = n2 * sin(r)
1 * sin(56.3) = 1.33 * sin(r)
r = 38.72 degrees
Now,
distance of googles = 1.8+ d*tan(r)
distance of googles = 1.8+ 3.1 * tan(38.72)
distance of googles = 4.3 mm
Answer:
The acceleration of the object is 
.
Explanation:
Given that,
Mass of the block, m = 2.6 kg
Spring constant of the spring, k = 126 N/m
At the instant when the displacement of the spring from its unstained length is 0.115 m. We need to find the acceleration of the object.
When the block is displaced, the force acting on the spring is equal to the force due to its motion. Such as :
a is acceleration of the object
So, the acceleration of the object is .
