Answer:
v_avg = 2.9 cm/s
Explanation:
The average velocity of the object is the sum of the distance of all its trajectories divided the time:
x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm
Then, x_all = 150cm + 140cm = 290cm
The average velocity is, for t = 100s
hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s
The second law of thermodynamics states that whenever energy changes occur, DISORDER always increases.
Answer:
The acceleration of the cart is 1.0 m\s^2 in the negative direction.
Explanation:
Using the equation of motion:
Vf^2 = Vi^2 + 2*a*x
2*a*x = Vf^2 - Vi^2
a = (Vf^2 - Vi^2)/ 2*x
Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.
Let x = Xf -Xi
Where Xf is the final position of the cart and Xi the initial position of the cart.
x = 12.5 - 0
x = 12.5
The cart comes to a stop before changing direction
Vf = 0 m/s
a = (0^2 - 5^2)/ 2*12.5
a = - 1 m/s^2
The cart is decelerating
Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.
Answer:
40
Explanation:
Mechanical advantage = effort arm / load arm
MA = 20 cm / 0.5 cm
MA = 40
The motion of the buoy consists of two independent motions on the horizontal and vertical axis.
On the horizontal axis, the motion of the buoy is a uniform motion with constant speed
. On the vertical axis, the motion of the buoy is a uniformly accelerated motion with constant acceleration
. The vertical position of the buoy at time t is given by
where h is the initial heigth of the buoy when it is released from the plane. At the time t=21 s, the buoy reaches the ground, so y(21 s)=0. If we substitute these two numbers inside the equation, we can find the value of h, the vertical displacement from the plane to the ocean:
