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3 years ago

What obstacles will you need to deal with to build railroad routes

Physics

1 answer:

polet [3.4K]3 years ago

4 0

Answer:

Construction of a railroad route is a strenuous task and one has to face several obstacles. 

Explanation:

Setting up the best possible plan for laying a rail route is a time-taking task. Several factors like economical nature of the plan, proper availability of building materials, minimum river crossings etc should be considered.

To build a railroad route the government has to get hold of the land through which the railroad route is planned. This is never an easy task. The people of the area have to be given adequate compensation and should be properly evacuated. The chances of people disapproving the plan are more.

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A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling

sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

$P_{1} = P_{2}$

$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

$v_{1_{i}}$ : is the initial velocity of the ball 1 = 6.20 m/s

$v_{2_{i}}$ : is the initial velocity of the ball 2 = 0 (it is at rest)

$v_{1_{f}}$ : is the final velocity of the ball 1 =?

$v_{2_{f}}$ : is the initial velocity of the ball 2 =?

$m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

$v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}}$ (1)

Now, by conservation of kinetic energy (since they collide elastically):

$\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}$

$m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}$ (2)

By entering equation (1) into (2) we have:

$m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}$

$0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}$

By solving the above equation for $v_{2_{f}}$ :

$v_{2_{f}} = 3.1 m/s$

Now, $v_{1_{f}}$ can be calculated with equation (1):

$v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s$

The minus sign of $v_{1_{f}}$ means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!

5 0

3 years ago

A NASA explorer spacecraft with a mass of 1,000 kg takes off in a positive direction from a stationary asteroid. If the velocity

Georgia [21]

Answer: 10000 kg

Explanation:

The momentum $p$ is given by the following equation:

$p=m.V$ (1)

Where:

$m$ is the mass of the object

$V$ is the velocity of the object

Now, in this case and according the conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1}+m_{2}u_{2}$ (2)

Where:

$m_{1}=1000kg$ is the mass of the spacecraft

$m_{2}$ is the mass of the asteroid

$v_{1}=0$ is the initial velocity of the spacecraft

$v_{2}=0$ is the initial velocity of the asteroid (because we are told the asteroid is stationary, as the spacracft is on the sateroid it remains stationary as well)

$u_{1}=250m/s$ is the final velocity of the spacecraft

$u_{2}=-25m/s$ is the final velocity of the asteroid

Rewritting (2):

$0=m_{1}u_{1}+m_{2}u_{2}$ (3)

$0=(1000kg)(250m/s)+m_{2}(-25m/s)$ (4)

Finding $m_{2}$ :

$m_{2}=10000kg$ This is the mass of the asteroid

3 0

3 years ago

Read 2 more answers

Describe how the life cycle of a low-mass star differs from the life cycle of a high-mass star. PLZZZZ ANSWER FAST

Alex Ar [27]

As the core collapses, the outer layers of the star are expelled. A planetary nebula is formed by the outer layers. The core remains as a white dwarf and eventually cools to become a black dwarf. that is what would happen to a star with a low mass like our sun also the life time of a star depends on it's mass. A larger mass star will colapse and turn into a black hole.

8 0

3 years ago

A particle moves along the x axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration 20.3

mixas84 [53]

Answer:

a) -2.34 m

b) -1.3 m/s

c) 0.056 m

d) 0.320 m/s

Explanation:

part a

Given:

s(0) = 0.27 m

v(0) = 0.14 m/s

a = -0.320 m/s^2

t = 4.50s

Using kinematic equation of motion for constant acceleration:

s (t) = s(0) + v(0)*t + 0.5*a*t^2

s ( 4.5 s ) = 0.27 + 0.14*4.5 + 0.5*-0.32*4.5^2

s(4.5) = -2.34 m

part b

Using kinematic equation of motion for constant acceleration:

v(t) = v(0) + a*t

v(4.5s) = (0.14) + (-0.32)(4.5) = -1.3 m/s

part c

Use equation for simple harmonic motion:

s(t) = A*cos(w*t)

v(t) = -A*w*sin (w*t)

a(t) = -A*(w)^2 * cos (w*t)

0.27 m = A*cos(w*t) .... Eq 1

0.14 m/s = - A*w*sin (w*t) .....Eq2

-0.320 = -A*(w)^2 * cos (w*t) .... Eq3

Solve the three equations above for A, w, and t

Divide 1 and 3:

w^2 = 0.32 / 0.27

w = 1.0887 rad / s

Divide 2 and 1:

w*tan(wt) = 0.14 / 0.27

tan(1.0887*t) = 0.476289

t = 0.4083 s

A = 0.27 / cos (1.0887*0.4083) = 0.3 m

Hence, the SHM is s(t) = 0.3*cos(1.0887*t)

s(4.5) = 0.3*cos(1.0887*4.5) = 0.056 m

part d

v(t) = - 0.32661 * sin (1.0887*t)

v(4.5) = - 0.32661 * sin (1.0887*4.5) = 0.320 m/s

3 0

3 years ago

A cylindrical container with a cross-sectional area of 85.2 cm 2 holds fluid having a density of 806 kg/m 3 . The surface of the

Anton [14]

Answer:

the answer would be B your welcome

Explanation:

5 0

3 years ago