Answer:
Vi = 32 [m/s]
Explanation:
In order to solve this problem we must use the following the two following kinematics equations.
![v_{f} =v_{i} - (a*t)\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20-%20%28a%2At%29%5C%5C)
The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.
where:
Vf = final velocity = 8[m/s]
Vi = initial velocity [m/s]
a = acceleration = [m/s^2]
t = time = 5 [s]
Now replacing:
8 = Vi - 5*a
Vi = (8 + 5*a)
As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.
![v_{f}^{2} = v_{i}^{2} - (2*a*d)](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%20%3D%20v_%7Bi%7D%5E%7B2%7D%20-%20%282%2Aa%2Ad%29)
where:
d = distance = 100[m]
(8^2) = (8 + 5*a)^2 - (2*a*100)
64 = (64 + 80*a + 25*a^2) - 200*a
0 = 80*a - 200*a + 25*a^2
0 = - 120*a + 25*a^2
0 = 25*a(a - 4.8)
therefore:
a = 0 or a = 4.8 [m/s^2]
We choose the value of 4.8 as the acceleration value, since the zero value would not apply.
Returning to the first equation:
8 = Vi - (4.8*5)
Vi = 32 [m/s]
Therefore, if the block moves from its position of maximum spring stretch to maximum spring compression in 0.25 s, the time required for a full cycle is twice as much; T = 0.5 s.
Answer:
Acceleration (a) = 40 m/s²
Explanation:
Given:
Initial velocity (u) = 6 m/s
Final velocity (v) = 4.4 m/s
Time taken (t) = 0.04sec
Find:
Acceleration (a) = ?
Computation:
We know that,
⇒ v = u + at
⇒ a = (v - u) / t
⇒ Acceleration (a) = (4.4 - 6) / 0.04
⇒ Acceleration (a) = (-1.6) / 0.04
Acceleration (a) = 40 m/s²
Answer:
The diagram represents two charges, q1 and q2, separated by a distance d. Which change would produce the greatest increase in the electrical force between the two charges? *
Explanation:
doubling charge q1, only
Opposite to the direction of the velocity which led it to its current position.
Explanation:
The direction of momentum when a vertically oscillating block comes to the rest momentarily will be opposite to the direction of the velocity that it has just followed to reach reach its current position.
The direction of change in momentum at the bottom will be upwards and at the top will be downwards.
The change in momentum is mathematically defined as:
![\Delta P=m.v_f-m.v_i](https://tex.z-dn.net/?f=%5CDelta%20P%3Dm.v_f-m.v_i)
where:
mass of the block
final velocity of the block
initial velocity of the block
When the block comes to rest it is due to the result of continuously decreasing velocity.