<span>Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get.......
17.15 m/s</span>
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
(a) A = 1 mm
(b)
(c)
(c) The formula for the maximum acceleration is given by
[tex]a_{max}=606.4 m/s^{2}/tex]
Answer:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
b) Determine the magnitude and direction of current in wire 3
Explanation:
a)
position of wire = 50 - 1.2
= 48.8cm
b)
Direction ⇒ downward
I am going to assume 2.1 metres per second and that we're rounding acceleration due to gravity to -10 metres per second squared. At the highest point, velocity is going to be 0. v= intial velocity + acceleration*time, sub in 0 for velocity, 2.1 for initial velocity and -10 for acceleration to get 0= 2.1-10t. Now solve for t. t=0.21 seconds.