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Kryger [21]
3 years ago
13

f two objects travel through space along two different curves, it's often important to know whether they will collide. (Will a m

issile hit its moving target? Will two aircraft collide?) The curves might intersect, but we need to know whether the objects are in the same position at the same time. Suppose the trajectories of two particles are given by the vector functions r1(t) = t2, 13t − 36, t2 r2(t) = 7t − 12, t2, 5t − 4
Physics
1 answer:
Pavel [41]3 years ago
3 0

Answer:

The two objects will collide with the same position vector for all three components at exactly t = 4 s

Explanation:

For two particles starting out at the same time to collide, their position Vector's at the time of collision must be exactly the same.

So, at the collision point, position vector of object 1 is equated to that of object 2.

r₁ = (t², 13t-36, t²)

r₂ = (7t-12, t², 5t-4)

At he point of collision

t² = 7t - 12

t² - 7t + 12 = 0

t² - 4t - 3t + 12 = 0

t(t - 4) - 3(t - 4) = 0

t = 3s or t = 4s

13t - 36 = t²

t² - 13t + 36 = 0

t² - 4t - 9t + 36 = 0

t(t - 4) - 9(t - 4) = 0

t = 9s or 4s

t² = 5t - 4

t² - 5t + 4 = 0

t² - 4t - t + 4 = 0

t(t - 4) - 1(t - 4) = 0

t = 1s or t = 4s

The three components intersect at other times, but at t = 4s, they all intersect at the same time! Meaning that, at this point the two objects are at the same place with the same position vector at that time.

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Inga [223]
Hope this helps, have a great day ahead!

7 0
3 years ago
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A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l
artcher [175]

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, \lambda_1=8\times 10^{-6}\ C

linear charge density of the other charge distribution, \lambda_2=-6\times 10^{-6}\ C

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

E=\frac{\lambda}{2\pi.r.\epsilon_0}

where:

\lambda= linear charge density

r = radial distance from the center of wire

\epsilon_0= permittivity of free space

Therefore,

E_1=E_2

\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}

\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}

\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}

x=0.0629\ m

∴at y=6.29 cm the charge of the two distribution will be equal.

9 0
3 years ago
A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is
ollegr [7]
<span>F = ma
</span>Ff = μ*Fn
<span>Fn = Fw
</span>Fw = mg 

<span>So we have: </span>

<span>Ff = μmg </span>

<span>And </span>

<span>Ff = ma </span>

<span>So... </span>

<span>μmg = ma </span><span> </span>

<span>μg = a </span>

<span>And we can solve for the acceleration: </span>

<span>(0.15)(9.81 m/s²) = a </span>

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8 0
3 years ago
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A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The
Goshia [24]

Answer:

0.629\ \text{rad/s}^2 counterclockwise

9.98\ \text{s}

Explanation:

r_1 = Small drive wheel radius = 2.2 cm

\alpha_1 = Angular acceleration of the small drive wheel = 8\ \text{rad/s}^2

r_2 = Radius of pottery wheel = 28 cm

\alpha_2 = Angular acceleration of pottery wheel

As the linear acceleration of the system is conserved we have

r_1\alpha_1=r_2\alpha_2\\\Rightarrow \alpha_2=\dfrac{r_1\alpha_1}{r_2}\\\Rightarrow \alpha_2=\dfrac{2.2\times 8}{28}\\\Rightarrow \alpha_2=0.629\ \text{rad/s}^2

The angular acceleration of the pottery wheel is 0.629\ \text{rad/s}^2.

The rubber drive wheel is rotating in clockwise direction so the pottery wheel will rotate counterclockwise.

\omega_i = Initial angular velocity = 0

\omega_f = Final angular velocity = 60\ \text{rpm}\times \dfrac{2\pi}{60}=6.28\ \text{rad/s}

t = Time taken

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_2t\\\Rightarrow t=\dfrac{\omega_f-\omega_i}{\alpha_2}\\\Rightarrow t=\dfrac{6.28-0}{0.629}\\\Rightarrow t=9.98\ \text{s}

The time it takes the pottery wheel to reach the required speed is 9.98\ \text{s}

4 0
3 years ago
A fish is able to jump vertically out of the water with a speed of 4.45 m/s. What is the speed of the fish as it passes a point
rjkz [21]

Answer:

Explanation:

given

initial velocity u = 4.45m/s

Height = 0.6m

g = 9.8m/s²

Required

final velocity v

Using the equation of motion;

v² = u²-2gH (upward motion of the fish makes g to be negative)

v² = 4.45²-2(9.8)(0.6)

v² = 19.8025-11.76

v² = 8.0425

v = 2.84 m/s

Hence the speed of the fish as it passes a point 0.6 m above the water is 2.84m/s

To get the time, we will use the formula

v = u - gt

2.84 = 4.45 - 9.8t

2.84-4.45 = -9.8t

-1.61 = -9.8t

t = 1.61/9.8

t = 0.164secs

Hence the time taken is 0.164secs

8 0
3 years ago
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