We are given the mass of an <span>aluminum sculpture which is 145 kg and a horizontal force equal to 668 Newtons. The coefficient of friction can be determined by dividing the horizontal force by the weight of the object. In this case, 668 N / 145 * 9.8 equal to coeff of friction of 0.47</span>
Answer:
t = 16.94 s
Explanation:
t is the time passes before police catch the speeder
speed of speeder Vo = V = 23.3 m/s
T = t
Police Info
Vo = 0 m/s
a = 2.75 m/s^2
t = t
Now,
displacement of the police car = displacement of the speeder.
x_{police} = Vo *t + 1/2 at^2
since Vo = 0
x police = 1/2 at^2
x police = 1/2 (2.75)(t)^2
Now the displacement of speeder is
x_{speeder} = Vt
x_{speeder} = 23.3 t
x_{speeder} = x_{police}
23.3 t = 1/2 * 2.75 t^2
23.3 t = 1.375 t^2
t = 23.3\1.375
t = 16.94
t = 16.94 s
Answer:
0.339 kgm²
Explanation:
We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.
Since T = 2π√(I/mgh), making I subject of the formula, we have
I = mghT²/4π²
Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.
So, T = 241 s/113 = 2.133 s
So, Substituting the values of the variables into I, we have
I = mghT²/4π²
I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²
I = 15.63/4π² kgm²
I = 0.396 kgm²
Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis
I' = I - mh²
I' = 0.396 kgm² - 2.15 kg × (0.163 m)²
I' = 0.396 kgm² - 0.057 kgm²
I' = 0.339 kgm²
Answer:
A
Explanation:
The figure shows the electric field produced by a spherical charge distribution - this is a radial field, whose strength decreases as the inverse of the square of the distance from the centre of the charge:
More precisely, the strength of the field at a distance r from the centre of the sphere is
where k is the Coulomb's constant and Q is the charge on the sphere.
From the equation, we see that the field strength decreases as we move away from the sphere: therefore, the strength is maximum for the point closest to the sphere, which is point A.
This can also be seen from the density of field lines: in fact, the closer the field lines, the stronger the field. Point A is the point where the lines have highest density, therefore it is also the point where the field is strongest.
