Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second particle is at (−7 + 2t, −6 + 2t, −1 + t). (a) (5 Points) Do the paths of the two particles cross? If so, where?

Answers

Answer 1

Answer:

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the paths meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

Related Questions

Power P is the rate at which energy E is consumed per unit time. Ornithologists have found that the power consumed by a certain pigeon flying at velocity v m/s is described well by the function P(v)=16v−1+10−3v3 J/s. Assume that the pigeon can store 5×104 J of usable energy as body fat. Find the velocity vp min that minimizes power consumption. (Use decimal notation. Give your answer to two decimal places.)
A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?
Explain what happent to the pressure exerted by an object when the area over which it is exerted:a) increase b) decrease
A ball with radius .15 m is rolling with an angular velocity of 6.5 rad/s. (a) What linear distance will the ball roll in 5.0 seconds? The ball then slows down with a linear acceleration of -1.2 rad/s2. (B) How fast will it be rolling rad/s after .65 second? (C) If a piece of gum is stuck on the outer part of the ball, what is its linear speed? (4.9, 5.72, .858)
A block of ice with mass 2.00 kg slides 0.775 m down an inclined plane that slopes downward at an angle of 31.8 ∘ below the horizontal. A) If the block of ice starts from rest, what is its final speed? You can ignore friction.

A diverging lens has a focal length of -30.0 cm. An object is placed 18.0 cm in front of this lens.(a) Calculate the image distance.

(b) Calculate the magnification.

Answers

Answer:

A) Calculate the distance

An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Determine the speed at this time and the maximum height at which it reaches.

Answers

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2* 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+(at^2)/(2)$

$s=0+(2* 6^2)/(2)$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=(144)/(2* 9.8)=7.34 m$

$s+s_0=36+7.34=43.34 m$

Which describes the motion of the box based on the resulting free-body diagram?1. It is moving up with a net force of 20 N.
2. It is moving to the right with a net force of 10 N.
3. It is in dynamic equilibrium with a net force of 0 N.
4. It is in static equilibrium with a net force of 0 N.

Answers

The statement "It is in dynamic equilibrium with a net force of 0 N" describes the motion of the box based on the resulting free-body diagram. (option 3)

What is a free-body diagram?

A free-body diagram is a diagram that shows all the forces acting on an object. If the net force on an object is zero, then the object is in equilibrium. This means that the object is not accelerating and is either at rest or moving with constant velocity.

In the case of the box in the free-body diagram, there are two forces acting on it: the force of gravity and the force of the table pushing up on the box. The force of gravity is pulling the box down, but the force of the table is pushing the box up.

These two forces are equal in magnitude and opposite in direction, so they cancel each other out. This means that the net force on the box is zero and the box is in dynamic equilibrium.

Learn about free-body diagram here brainly.com/question/10148657

#SPJ3

Answer:

4. It is in static equilibrium with a net force of 0 N.

Explanation:

Just got it right :)

A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?

Answers

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

$\mu mg = (mv^2)/(R)$

here we have

$\mu g = (v^2)/(R)$

now we know that

$v = √(\mu Rg)$

here we have

$\mu = 0.600$

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

$v = √((0.600)(35)(9.81))$

$v = 14.35 m/s$

A 372-g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is 17.6 m/s2 , and its maximum speed is 1.75 m/s. a)Determine the angular frequency. b)Determine the amplitude. c)Determine the spring constant.

Answers

Answer with Explanation:

We are given that

Mass , m=372 g= $(372)/(1000)=0.372 Kg$

1 kg=1000g

Maximum acceleration, a= $17.6 m/s^2$

Maximum speed ,v=1.75 m/s

a.We know that

Maximum acceleration, a= $A\omega^2$

Maximum speed, v= $\omega A$

$17.6=A\omega^2$

$1.75=A\omega$

$(17.6)/(1.75)=(A\omega^2)/(A\omega)=\omega$

Angular frequency, $\omega=10.06 rad/s$

b.Substitute the value of angular frequency

$1.75=A(10.06)$

$A=(1.75)/(10.06)=0.17 m$

Hence, the amplitude=0.17 m

c.Spring constant,k= $m\omega^2$

Using the formula

$k=0.372* (10.06)^2$

Hence, the spring constant,k=37.6 N/m

An object has an average acceleration of +6.07 m/s2 for 0.250 s . At the end of this time the object's velocity is +9.64 m/s .What was the object's initial velocity?