The applied force is different for the two cases
The case A with a greater force involves the greatest momentum change
The case A involves the greatest force.
<h3>What is collision?</h3>
- This is the head-on impact between two object moving in opposite or same direction.
The initial momentum of the two ball is the same.
P = mv
where;
- m is the mass of each
- v is the initial velocity of each ball
Since the force applied by the arm is different, the final velocity of the balls before stopping will be different.
Thus, the final momentum of each ball will be different
The impulse experienced by each ball is different since impulse is the change in momentum of the balls.
J = ΔP
The force applied by the rigid arm is greater than the force applied by the relaxed arm because the force applied by the rigid arm will cause the ball to be brought to rest faster.
Thus, we can conclude the following;
- The applied force is different for the two cases
- The case A with a greater force involves the greatest momentum change
- The case A involves the greatest force.
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Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
It is Real,Virtual,The Same Size, Inverted
It will be traveling in the reverse direction it was originally going at 15.2 m/s
4200 N is the tension in the cable that pulls the elevator upwards.
The correct option is A.
<h3>What does tension ?</h3>
Tension is the force that is sent through a rope, thread, or wire whenever two opposing forces pull on it. Along the whole length of the wire, the tensile stress pulls equally on all objects at the ends. Every physical object that comes into contact with that other one exerts force on it.
<h3>Briefing:</h3>
We employ the following formula to determine the cable's tension.
Formula:
T = mg+ma............ Equation 1
Where:
T is the cable's tension.
M = Mass of the elevator and the Joey
Accelerating with a
g = Gravitational acceleration
Considering the query,
Given:
m = (300+60) = 360 kg
a = 2 m/s²
g = 9.8 m/s²
Substitute these values into equation 2
T = (360×9.8)+(360×2)
T = 3528+720
T = 4248 N
T ≈ 4200 to the nearest hundred.
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