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2 years ago

A) A 12 kg object has a velocity of 37.5 m/s. What is its momentum?

Physics

1 answer:

Mkey [24]2 years ago

4 0

Answer:

The answer would be 450 m kg/s

Explanation/ Explanation / Example:

Provided an object traveled 500 meters in 3 minutes , to calculate the average velocity you should take the following steps: Change minutes into seconds (so that the final result would be in meters per second). 3 minutes = 3 * 60 = 180 seconds , Divide the distance by time: velocity = 500 / 180 = 2.77 m/s .

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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

What is the density of this liquid that has a mass of 300g and volume 600ml.

8_murik_8 [283]

Answer:

0.5

Explanation:

D = M / V

You divide 300g by 600ml and you get 0.5

(✿◠‿◠)

8 0

3 years ago

Read 2 more answers

An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 14.0 m if her initial speed is 2.20 m/

pochemuha

We have the equation of motion

$v^2 = u^2+2as$ , where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.

In this case initial velocity = 2.20 m/s, final velocity = 0 m/s, displacement = 14 m

On substitution we will get 0 = $2.20^2+2*a*14$

On solving we can find the acceleration value as -0.173 $m/s^2$

So free fall acceleration = 0.173 $m/s^2$

3 0

3 years ago

If the distance between two asteroids is halved, the gravitational force they exert on each other will

mamaluj [8]

Answer:

e) Be four times greater

Explanation:

Here we have to use Newton's gravitational law that relates the gravitational force between two objects with their masses ( $m_{1}$ & $m_{2}$ ) and the distance between them ( $r$ ) in the next way:

$F=G\frac{m_{1}m_{2}}{r^{2}}$ (2)

Now if distance between asteroids is halved:

$F_{2}=G\frac{m_{1}m_{2}}{(\frac{r}{2})^{2}}$

$F_{2}=G\frac{m_{1}m_{2}}{\frac{r^{2}}{4}}$

$F_{2}=G\frac{4m_{1}m_{2}}{r^{2}}=4G\frac{m_{1}m_{2}}{r^{2}}$

Note that $G\frac{m_{1}m_{2}}{r^{2}}$ because (1) is F so:

$F_{2}=4F$

It's four times greater!

3 0

3 years ago

Technician A says that shop air pressure usually ranges from 100–150 psi. Technician B says shop air pressure is around 300 psi.

mr Goodwill [35]

Answer:

technician A.

Explanation:

Commonly, air pressure rating of standard shop air compressor's is between the range of 100 psi and 150 psi. And over the years since, the plant pressure creep's up , the level of pressure increases from 100 psi to at max 150 psi but not more than that. Therefore, in the given question technician A is correct and technician B is incorrect.

5 0

3 years ago