\Delta L= \alpha L_0 (T_f-T_i)
= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
So your answer is B.
Answer:
L/2
Explanation:
Neglect any air or other resistant, for the ball can wrap its string around the bar, it must rotate a full circle around the bar. This means the ball should be able to swing to the top position where it's directly above the bar. By the law of energy conservation, this happens when the ball is at the same level as where it's previously released vertically. It means the swinging radius around the bar must be at least half of the string length.
So the distance d between the bar and the pivot should be at least L/2
This is simple as power in watts is equal to joules per second so we can do 1500 joules divided by 30 seconds which equals 50 watts
Actually what the problem meant about the westward
component of the ball’s displacement is the horizontal component of the
displacement. To help us better understand the problem, I attached a figure of
the situation.
We can see from the figure that to solve for the value of
the horizontal component, we have to make use of the sin function. That is:
sin θ = side opposite to the angle / hypotenuse of the
triangle
sin 42 = x / 40 m
x = (40 m) sin 42
x = 26.77 m
Therefore the ball has a westward
displacement of about 26.77 m