Momentum of the 2 kg mass moving with velocity 10 m/s is *

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

2 years ago

Convert 0.0375 kg to g

leonid [27]

Answer: 37.5 g

Explanation: you multiply 0.0375 by 1000 which equals 37.5

8 0

3 years ago

Read 2 more answers

If you are looking for a material that is a good conductor, you will look for a material that will (2 points)

Nataly_w [17]

B)<span>Readily accept electron flow.</span>

7 0

3 years ago

Read 2 more answers

As part of the Antarctic Treaty System, all facilities on Antarctica are inspected. What is one reason there's a need for these

Alinara [238K]

Answer:

D. To ensure common resources are not overused

Explanation:

The article of the antarctic treaty VII provides that all the consultative parties have a right to designate the observers and to undertake the inspection as of the protocol 14 of that provides the protection of the environment and to safeguard the ecological system.

Since 1963, about 9 inspections have been conducted by the teams by visiting the stations, facilities, the vessels, and the protected areas across a wide area of East Antarctica, The most recent Antarctica expedition treaty inspection was done in Dec 2016.

5 0

2 years ago

Create the following matrix H:

sladkih [1.3K]

Answer:

a) {[1.25 1.5 1.75 2.5 2.75]

[35 30 25 20 15] }

b) {[1.5 2 40]

[1.75 3 35]

[2.25 2 25]

[2.75 4 15]}

Explanation:

Matrix H: {[1.25 1.5 1.75 2 2.25 2.5 2.75]

[1 2 3 1 2 3 4]

[45 40 35 30 25 20 15]}

Its always important to get the dimensions of your matrix right. "Roman Columns" is the mental heuristic I use since a matrix is defined by its rows first and then its column such that a 2 X 5 matrix has 2 rows and 5 columns.

Next, it helps in the beginning to think of a matrix as a grid, labeling your rows with letters (A, B, C, ...) and your columns with numbers (1, 2, 3, ...).

For question a, we just want to take the elements A1, A2, A3, A6 and A7 from matrix H and make that the first row of matrix G. And then we will take the elements B3, B4, B5, B6 and B7 from matrix H as our second row in matrix G.

For question b, we will be taking columns from matrix H and making them rows in our matrix K. The second column of H looks like this:

{[1.5]

[2]

[40]}

Transposing this column will make our first row of K look like this:

{[1.5 2 40]}

Repeating for columns 3, 5 and 7 will give us the final matrix K as seen above.

4 0

3 years ago