How many sides and points do a hexagon have

(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50

Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, $p=m_Av_A=m_Bv_B=m_Cv_C$ , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as $m_h$ , and the mass of the first and second coals as $m_1$ and $m_2$ respectively

We start with the transition between parts A and B, so we have:

$m_Av_A=m_Bv_B$

Which means

$m_hv_A=(m_h+m_1)v_B$

And since we want the mass of the first coal thrown ( $m_1$ ) we do:

$m_hv_A=m_hv_B+m_1v_B$

$m_hv_A-m_hv_B=m_1v_B$

$m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}$

Substituting values we obtain

$m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg$

For the transition between parts B and C, we can write:

$m_Bv_B=m_Cv_C$

Which means

$(m_h+m_1)v_B=(m_h+m_1+m_2)v_C$

Since we want the new final speed of the car ( $v_C$ ) we do:

$v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}$

Substituting values we obtain

$v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s$

5 0

2 years ago

4. A trolley of mass 2kg rests next to a trolley of mass 3 kg on a flat

nydimaria [60]

Answer:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. The total momentum of the trolleys after separation is zero

c. The momentum of the 2 kg trolley after separation is 12 kg·m/s

d. The momentum of the 3 kg trolley is -12 kg·m/s

e. The velocity of the 3 kg trolley = -4 m/s

Explanation:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0

c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s

d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s

e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley

∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s

The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s

3 0

2 years ago

What force acts in the opposite direction to an object moving through the air?

evablogger [386]

Air resitances also know as drag is a force that is caused by air the force acts in oppsite directions to an object moving through the air ..

HOPE THIS HELP YOU ..

5 0

2 years ago

A sound wave will rebound when it encounters what?

Over [174]

The sound wave will rebound when it is a barrier, hope this helps

8 0

3 years ago

Read 2 more answers

1. You are about to make a left turn. You must signal continuously during the last

Bingel [31]

Answer:

B

Explanation:

5 0

3 years ago