Explain how water changes state using physical changes.

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

So

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago

What provides the force on the person in the passenger seat?

podryga [215]

The forces that make a passenger speed up, slow down, or
turn a curve are the same forces that have the same effect
on the driver and anybody else in the car.

-- Speeding up . . .

the back of the seat
    friction between the car seat and the seat of your pants

-- Slowing down . . .

the seat belt
      friction between the car seat and the seat of your pants

-- Turning away from a straight line . . .

      the seat belt
friction between the car seat and the seat of your pants
the door, or whatever or whomever you're leaning against

6 0

3 years ago

The blades in a blender rotate at a rate of 7700 rpm. when the motor is turned off during operation, the blades slow to rest in

Tpy6a [65]

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (speed at the end) - (speed at the beginning)

For this fan, speed at the end = 7700 rpm, speed at the end = 0 .

Change in angular speed = -7700 rpm

Angular acceleration = (-7700 rpm) / (2.5 sec)

<em>Angular acceleration = -3,080 rev per minute / sec</em>

That's a perfectly good and true answer to the question, but the units are ugly. We really need to fix the units, and convert them into something prettier before we hand in this assignment.

1 rev = 2π radians, and

1 minute = 60 seconds .

So

Angular acceleration =

(-3,080 rev/min-sec) · (2π rad/rev) · (1 min/60 sec)

AngAccel = (-3,080 · 2π · 1 / 60) · (rev·rad·min / min·sec·rev·sec)

AngAccel = ( -102 and 2/3 · π) · (rad/s²)

<em>AngAccel = -322.5 radian/s²</em>

7 0

3 years ago

Light waves have some similarities with water and sound waves, but they are not exactly the same. Describe all the differences y

makkiz [27]

<u>Answer:</u>

<h2>All the waves are pertubations that propagate (transport) energy.</h2><h2></h2>

Nevertheless, they have some differences:

1. Light waves are<u> electromagnetic waves</u>, while sound and water waves are <u>mechanical waves</u>, this is the first and principal difference.

2. Electromagnetic waves can<u> propagate in vacuum</u> (they do not need a medium or material), but mechanical waves obligatory need a material to propagate

3. Light waves are always <u>transversal waves</u>, this means <u>the oscillatory movement is in a direction that is perpendicular to the propagation</u>; but mechanical waves may be both: <u>longitudinal waves</u> (the oscillation occurs in the same direction as the propagation) or transversal waves.

4. Electromagnetic waves propagates at a <u>constant velocity</u> (Light velocity) while the velocity of mechanical waves will depend on the type of wave and the <u>density</u> of the medium or material.

5. <u>Mechanical waves</u> are characterized by the regular variation of a single magnitude, while <u>electromagnetic waves</u> are characterized by the variation of two magnitudes: the electric field and the magnetic field

6. <u>Water waves</u> are 2-dimensional waves, while the <u>light and the sound</u> are tridimensional spherical waves

7. Light waves <u>transports energy in the form of </u><u>radiation</u>, while mechanical waves t<u>ransport energy with </u><u>material</u>

3 0

3 years ago

We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth

Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

p_f = (m + M) v

the moment is preserved

p₀ = p_f

m v₀ = (m + M) v

v = $\frac{m}{m+M} \ v_o$

for when n balls have collided

v = $\frac{m}{n \ m + M}$ v₀

Now let's analyze the case of the bouncing ball (elastic)

initial instant

p₀ = m v₀

final moment

p_f = m v_{1f} + M v_{2f}

p₀ = p_f

m v₀ = m v_{1f} + M v_{2f}

m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

K₀ = K_f

½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

v₁ = v_{1f} v₂ = v_{2f}

m (v₀² - v₁²) = M v₂²

let's use the identity

(a² - b²) = (a + b) (a-b)

we write our equations

m (v₀ - v₁) = M v₂ (1)

m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

m (v₀ - v₁) = M (v₀ + v₁)

v₀ (m -M) = (m + M) v₁

v₁ = $\frac{m-M}{m + M}$ v₀

we substitute in equation 1 to find v₂

$\frac{M}{m}$ v₂ = v₀ - $\frac{m-M}{m+M}$ v₀

v₂ = $\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o$

v₂ = $\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o$

v₂ = $\frac{2m}{m +M} \ v_o$

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

8 0

2 years ago