Answer:
v = 15 m / s
Explanation:
In this exercise we are given the position function
x = 5 t²
and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval
let's look for the displacements
t = 0 x₀ = 0 m
t = 3
= 5 3 2
x_{f} = 45 m
we substitute

v = 15 m / s
Answer:
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Explanation:
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For this case, the first thing we must do is define a reference system.
Suppose that the positive direction of the reference system is upward.
We have that the sum of forces in the vertical axis is given by:
Fy = Fp - Fg
Substituting values:
Fy = 5500 - 6000
Fy = - 500
The negative sign means that the direction of the force with respect to the defined coordinate system is downward.
Answer:
The net force is:
↓ 500N
The answer is letter B. XD
Answer:
Vf=3
Explanation:
you must first write your data
data before impact
M1=1000 M2=5000
V1=0 m/s V2 =0m/s
data after impact
M1=1000 M2=5000
V1=15m/s V2=?
M1V1 +M2V2=M1V1 +M2V2f
(1000)(0)+(5000)(0)=(1000)(15)+(5000)Vf
0=15000+5000Vf
- 15000÷5000=5000Vf÷5000
Vf= -3
Vf =3