2. The ability to keep<br> upright posture while standing or moving

Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the

Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = ( $\frac{1}{5}$ × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂ = 14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0

3 years ago

Compare and contrast the theories about the origin of the universe

Vaselesa [24]

Answer:

The best-supported theory of our universe's origin centers on an event known as the big bang. This theory was born of the observation that other galaxies are moving away from our own at great speed in all directions, as if they had all been propelled by an ancient explosive force.

Explanation:

hope this helps tho i don't quite know what you mean

7 0

3 years ago

A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

cupoosta [38]

Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively: