Since the collision is inelastic, both vehicle will move with a common velocity of 2.5 m/s in south east direction after collision.
<h3>
Conservation of Linear Momentum</h3>
It states that, the sum of the momentum before collision is equal to the sum of the momentum after collision.
Given that a 7500 kg truck traveling at 5.0 m/s east collides with a 1500 kg car moving at 20 m/s in a direction 30° south of west. After collision, the two vehicles remain tangled together.
The given parameters are;
The formula to find the speed will be
M1U1 - M2U2 = (M1 + M2)V
7500 × 5 - 1500 × 10 = (7500 + 1500)V
37500 - 15000 = 9000V
22500 = 9000V
V = 22500/9000
V = 2.5 m/s
Therefore, the common speed will be 2.5 m/s and the direction of the car will be toward the south east.
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Answer:
It is easier to stop the bicycle moving at a lower velocity because it will require a <em>smaller force</em> to stop it when compared to a bicycle with a higher velocity that needs a<em> bigger force.</em>
Explanation:
The question above is related to "Newton's Law of Motion." According to the <em>Third Law of Motion</em>, whenever an object exerts a force on another object <em>(action force)</em>, an equal force is exerted against it. This force is of the same magnitude but opposite direction.
When it comes to moving bicycles, the force that stops their movement is called "friction." Applying the law of motion, the higher the speed, the higher the force<em> </em>that is needed to stop it while the lower the speed, the lower the force<em> </em>that is needed to stop it.
We can’t exactly measure how fast speed is traveling .
S- sexually T- transmitted D- disease
These are usually transmitted by some sort of sexual inter-coarse .
some examples of STD's are HIV,chlamydia,genital herpes,&gonorrhea . These are not the only ones there are many other types of STDs
The value of t for which the particle is at rest is : t = 2
<u>Given data:</u>
X = t³ - 3t²
Y = 2t³ - 3t² - 12t
<h3>Determine the values of t for which the particle is at rest </h3>
<u>First step : Determine the first derivative of eac equations</u>
dx = 3t² - 6t
dy = 6t² - 6t - 12
<u>Next step : </u><u>determine</u><u> the value of slope ( dy/dx ) = o </u>
dy / dx = ( 6 (t² - t - 2) ) / ( 3t ( t - 2) )
Therefore
dy / dx = ( 2 ( t + 1 ) (t - 2) ) / ( t ( t -2) )
= 0
Therefore at ; t = -1 and 2 the particle is at rest
Hence we can conclude that Neglecting the negative value the particle will be at rest when t = 2
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<u>Attached below is the complete question</u>
<em>The position of a particle moving in the xy-plane is given by the parametric equations x = t3 - 3t2 and y = 2t3 - 3t2 - 12t. For what values of t is the particle at rest?</em>