Answer:
L2 = 1.1994 m
the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m
Explanation:
Given;
Initial length L1 = 1.2m
Initial temperature T1 = 27°C
Final temperature T2 = 0.0°C
Linear expansion coefficient of brass x = 1.9 × 10^-5 /°C
The change i length ∆L;
∆L = L2 - L1
L2 = L1 + ∆L ...........1
∆L = xL1(∆T)
∆L = xL1(T2 - T1) ......2
Substituting the given values into equation 2;
∆L = 1.9 × 10^-5 /°C × 1.2m × (0 - 27)
∆L = 1.9 × 10^-5 /°C × 1.2m × (- 27)
∆L = -6.156 × 10^-4 m
From equation 1;
L2 = L1 + ∆L
Substituting the values;
L2 = 1.2 m + (- 6.156 × 10^-4 m)
L2 = 1.2 m - 6.156 × 10^-4 m
L2 = 1.1993844 m
L2 = 1.1994 m
the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m
At position of maximum height we know that the vertical component of its velocity will become zero
so the object will have only horizontal component of velocity
so at that instant the motion of object is along x direction
while if we check the acceleration of object then it is due to gravity
so the acceleration of object is vertically downwards
so it is along y axis
so here these two physical quantities are perpendicular to each other
so correct answer would be
<em>C)At the maximum height, the velocity and acceleration vectors are perpendicular to each other. </em>
Answer:
The initial velocity of the ball is 28.714 m/s
Explanation:
Given;
time of flight of the ball, t = 2.93 s
acceleration due to gravity, g = 9.8 m/s²
initial velocity of the ball, u = ?
The initial velocity of the ball is given by;
v = u + (-g)t
where;
v is the final speed of the ball at the given time, = 0
g is negative because of upward motion
0 = u -gt
u = gt
u = (9.8 x 2.93)
u = 28.714 m/s
Therefore, the initial velocity of the ball is 28.714 m/s
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.
1) Equilibrium of forces:
where
is the weight of the person
is the weight of the scaffold
Re-arranging, we can write the equation as
(1)
2) Equilibrium of torques:
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using
and replacing T1 with (1), we find
from which we find
And then, substituting T2 into (1), we find
