Active transport is a process that uses energy to move materials through a

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

I hope it helps you!

3 0

3 years ago

A bat hits a ball; which has the greater acceleration, the bat or the ball?

MaRussiya [10]

The ball because the Kinetic Energy transfers from the bat to the ball, increasing the movement and acceleration of the ball because of the Kinetic Energy transferred from the origin force (The bat)

7 0

3 years ago

Read 2 more answers

If you put 120 volts of electricity through a pickle, the pickle will smoke and start glowing orange-yellow. The light is emitte

Alona [7]

Answer:

2.11eV

Explanation:

We know that speed of light is it's wavelength times frequency.

$\therefore f=v/\lambda\\=(3\times10^8m/s)/(589mm\times1m/1\times 10^9nm)\\=5.09\times10^1^4s^-1 \ or \ 5.09\times10^1^4Hz$

Planck's constant is $6.626\times 10^3^4Js$

The energy gap is calculated by multyplying the light's frequency by planck's constant:

$E_c=5.09\times10^1^4s^-^1\times 6.626\times10^-^3^4Js\\\\=3.37\times 10^-^1^9J \ \ \ \ \ \ #1eV=1.06\times 10^-^1^9J\\\\=2.11eV$

Hence, the energy gap is 2.11eV

4 0

3 years ago

You are on a rollercoaster going around a curve. Your speed is a constant 40 miles per hour. Are you accelerating?

vodomira [7]

No, you are at a constant rate which means that you are always at 40mph

8 0

3 years ago

Let's talk skateboard velocity how much mean when you get my going down a 3 x 3" metal ramp. No posers are allowed to answer th

denpristay [2]

Answer:

quarte embri Oslo

Explanation:

8 0

2 years ago

Read 2 more answers