Answer: 6s
Explanation:
Vs=32m/s speed at beginning of slowing down
Vf=0m/s stop speed
a= -6 m/s² acceleration
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Use equation for acceleration :
a=(Vf-Vs)/t
a*t=Vf-Vs
t=(Vf-Vs)/a
t=(0-36)/-6
t=-36/-6
t=6 s
Odometer: tells you the distance traveled by vehicle since it was new (or when last reset)
Speedometer: tells you the velocity of the vehicle at that moment.
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
Answer:
I'll try to help which grade are you?
The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.
The given parameters;
<em>Mass of the first object, m1 = 1 kg</em>
<em>Mass of the second object, m2 = 5 kg</em>
The final velocity of the objects during the downward motion is calculated as follows;
The time of motion of the object from the given height is calculated as;
The time of motion of each object is independent of mass of the object.
Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.
Learn more about time of motion here: brainly.com/question/2364404