Let the angle be Θ (theta)
Let the mass of the crate be m.
a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.
Normal force (N) = mg CosΘ
μ (coefficient of static friction) = 0.29
Static friction = μN = μmg CosΘ
Now, along the ramp, the equation of net force will be:
mg SinΘ - μmg CosΘ = 0
mg SinΘ = μmg CosΘ
tan Θ = μ
tan Θ = 0.29
Θ = 16.17°
b) Let the acceleration be a.
Coefficient of kinetic friction = μ = 0.26
Now, the equation of net force will be:
mg sinΘ - μ mg CosΘ = ma
a = g SinΘ - μg CosΘ
Plugging the values
a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96
a = 2.7244 - 2.44608
a = 0.278 m/s^2
Hence, the acceleration is 0.278 m/s^2
Answer:
A) - 1.8 m/s
Explanation:
As we know that whole system is initially at rest and there is no external force on this system
So total momentum of the system must be conserved
so we will have
now plug in all data into above equation
so correct answer is
A) - 1.8 m/s
The answer would be 46.482 because you multiply 18.3 by 2.54 because for every inch you get 2.54 centimeters
Answer:
East component is: 18.64 m/s
Explanation:
If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:
East component = 32.5 m/s * sin(35) = 18.64 m/s
Explanation:
Given that,
Work done to stretch the spring, W = 130 J
Distance, x = 0.1 m
(a) We know that work done in stretching the spring is as follows :
(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m
So,
So, the new work is more than 130 J.
