As a river's slop increases, the power of the river to cause erosion usually _____

For each of the problems below, you will need to draw a graph to find the solution.

Gelneren [198K]

Answer:

Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Use standard gravity, a = 9.80665 m/s2, for equations involving the Earth's gravitational force as the acceleration rate of an object.

Explanation:

3 0

3 years ago

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

Suppose a soccer player kicks the ball from a distance 29 m toward the goal. find the initial speed of the ball if it just passe

Rudik [331]

The ball's vertical velocity at the time it just passes over the goal is 0 m/s. Its initial vertical velocity is unknown and we denote it by $v\sin39^\circ$ , where $v$ here is the ball's initial speed. Vertically, the only force acting on the ball is gravity, which attributes a downward acceleration of 9.8 m/s^2. We expect the maximum height achieved by the ball to be 2.4 m, so we can find the initial speed by solving

$\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(v\sin39^\circ\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm m)$

$\implies v=11\,\dfrac{\mathrm m}{\mathrm s}$

6 0

3 years ago

A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.64 10-2 kg/s. the density of the gasoline is 735

Irina18 [472]

Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m

Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²

Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s

The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s

Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.

7 0

3 years ago

Read 2 more answers

The force between two charged bodies is called ______________.

stepladder [879]

The Coulomb force between two or more charged bodies is the force between them due to Coulomb's law. If the particles are both positively or negatively charged, the force is repulsive; if they are of opposite charge, it is attractive. ... Like the gravitational force, the Coulomb force is an inverse square law.

6 0

4 years ago

Read 2 more answers

As a river's slop increases, the power of the river to cause erosion usually ______.