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2 years ago

Whích best explains parallel forces?

Physics

1 answer:

Alex787 [66]2 years ago

4 0

Answer:

Parallel forces lie on the same plane and they have lines of action that never intersect each other.

Explanation:

Parallel forces definition is - forces acting in parallel lines.

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In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight

algol13

Answer:

$v_{f} =25m/s$

Explanation:

Kinematics equation for constant acceleration:

$v_{f} =v_{o} + at=15+2*5=25m/s$

4 0

2 years ago

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T

zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f) The two trains avoid collision if the acceleration of the freight train is at least $0.35 m/s^2$

Explanation:

We can describe the position of the passenger train at time t with the equation

$x_p(t)=u_p t + \frac{1}{2}at^2$

where

$u_p = 25.0 m/s$ is the initial velocity of the passenger train

$a=-0.100 m/s^2$ is the deceleration of the train

On the other hand, the position of the freight train is given by

$x_f(t)=x_0 + v_f t$

where

$x_0=200 m$ is the initial position of the freight train

$v_f = 15.0 m/s$ is the constant velocity of the train

The collision occurs if the two trains meet, so

$x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0$

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

$x_f(t)=x_0 +v_ft$

And substituting t = 22.5 s, we find:

$x_f(22.5)=200+(15)(22.5)=537.5 m$

We can verify that the passenger train is at the same position at the time of the collision:

$x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m$

So, the two trains collide at x = 537.5 m.

In the graph in attachment, the position-time graph of each train is represented. We have:

The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

In order to avoid the collision, the freight train must have a initial position of

$x_0'$

such that the two trains never meet.

We said that the two trains meet if:

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t$

Re-arranging,

$\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0$

Substituting the values for the acceleration and the velocity,

$0.05t^2-10t+x_0'=0$

The solution of this equation is given by the formula

$t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}$

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

$10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m$

Therefore, the freight train must have a head start of 500 m.

In this case, we want to find the acceleration $a'$ of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t$

Re-arranging

$\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0$

Substituting,

$-0.5at^2-10t+200=0$

The solution to this equation is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}$

Again, the two trains never meet if the discriminant is negative, so

$10^2-4\cdot (-0.5a') \cdot (200)$

So, the deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

$x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2$

where $a_f$ is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2$

Re-arranging,

$\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0$

And the solution is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}$

Again, the two trains avoid collision if the discriminant is negative, so

$10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2$

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8 0

2 years ago

Describe how the design of a vacuum flask keeps the liquid inside hot ?

JulijaS [17]

The gap between the two flasks is partially evacuated of air creating a near vacuum which significantly reduces heat transfer by conduction or convection

3 0

3 years ago

Read 2 more answers

Earth is slightly closer to the Sun in January than in July. How does the area swept out by Earth's orbit around the Sun during

suter [353]

Answer: Option <em>a.</em>

Explanation:

Kepler's 2nd law of planetary motion states:

<em>A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.</em>

It tells us that it doesn't matter how far Earth is from the Sun, at equal times, the area swept out by Earth's orbit it's always the same independently from the position in the orbit.

3 0

3 years ago

How long does it take electrons to get from

OlgaM077 [116]

We need to find the time it takes an electron to move in the given circuit.

The time taken for electrons to reach the starting motor from the battery is 60.65 minutes.

I = Current = 134 A

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

A = Area = $38.9\ \text{mm}^2$

L = Length = 92.2 cm

$\rho$ = Density of copper = $8960\ \text{kg/m}^3$

M = Molar mass of copper = 63.5 g/mol

$n_v$ = Number of valence electrons of copper = 1

e = Charge of electron = $1.6\times 10^{-19}\ \text{C}$

Number of charge carriers per unit volume is given by

$n=\dfrac{\rho N_An_v}{M}\\\Rightarrow n=\dfrac{8960\times 6.022\times 10^{23}\times 1}{63.5\times 10^{-3}}\\\Rightarrow n=8.497\times 10^{28}\ \text{m}^{-3}$

Time taken is given by

$t=\dfrac{LAne}{I}\\\Rightarrow t=\dfrac{92.2\times 10^{-2}\times 38.9\times 10^{-6}\times 8.497\times 10^{28}\times 1.6\times 10^{-19}}{134}=3638.83\ \text{s}\\\Rightarrow t=\dfrac{3638.83}{60}=60.65\ \text{minutes}$

The time taken for electrons to reach the starting motor from the battery is 60.65 minutes.

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1 year ago