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iVinArrow [24]
3 years ago
9

Think about the various energy sources such as the sun, fossil fuels, water, wind, and nuclear power. What is the energy source

to power your home? Do you prefer this energy source? If not, which energy source would you prefer was used to power your house?
Physics
1 answer:
lakkis [162]3 years ago
8 0

Answer:

<em>The energy source that powers my home is gotten from burning of fossil fuels.</em>

<em>No! I do not like this energy source.</em>

<em>I personally would prefer solar source of electricity</em>

<em></em>

Explanation:

<em>Fossil fuels are fuels gotten from the decomposition of dead organisms over time, under intense heat and pressure</em>. They are usually found buried beneath the earth's crust where they have been formed and trapped.

Most electricity generating stations generate electricity by burning fossil fuels like natural gas and gasoline to generate electricity. <em>The problem with fossil fuels are the various part that they play in increasing carbon footprints in the atmosphere. The excess carbon in the atmosphere has been a major contributor to the global warming of planet Earth.  </em>

My preference for solar energy source is first due to its abundance unlike the fossil fuels that are already diminishing in storage beneath the earth's crust. Also, <em>solar energy is a clean source of energy that does not leave any damage on earth from its use</em>. It also promises to be a cheap source of power in the future with advances in solar technologies.

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Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s
anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of -5.2^{\circ}

D) The impulse exerted on C is 4.29 kg m/s at a direction of -5.2^{\circ}

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B

m_A = 0.600 kg is the mass of sphere A

u_A = 4.00 m/s is the initial velocity of the sphere A (taking the right as positive direction)

v_A is the final velocity of sphere A

m_B = 1.80 kg is the mass of sphere B

u_B = 2.00 m/s is the initial velocity of the sphere B

v_B = 3.00 m/s is the final velocity of the sphere B

Solving for vA:

v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s

The sign is positive, so the direction is to the right.

B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J

After the collision:

K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s

While the total momentum along the y-axis is zero:

p_y = 0

We can now write the equations of conservation of momentum along the two directions as follows:

p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy} (1)

We also know the components of the momentum of B after the collision:

p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s

So substituting into (1), we find the components of the momentum of C after the collision:

p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s

So the magnitude of the momentum of C is

p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s

And the direction is

\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}

D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

p_C = 0

While the final momentum is:

p'_C = 4.29 kg m/s

So the magnitude of the impulse exerted on C is

I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore -5.2^{\circ}.

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

F=Ma

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

brainly.com/question/6439920

brainly.com/question/2990238

brainly.com/question/7973509

brainly.com/question/6573742

#LearnwithBrainly

8 0
3 years ago
Technician A says that current is created in the stator windings of an alternator. Technician B says that the voltage regulator
drek231 [11]

Answer:

Technician A and Technician B both are right.

Explanation:

In an AC alternator,  there are two windings

1. Stator winding (stationary)

2. Rotor winding (rotating)

The current is induced in the stationary coils due to the magnetic field produced by the rotor. The DC suppy is provided to the rotor winding via slip rings and brushes and a voltage regulator precisely controls this supply to control the current flow through the rotor.

Therefore, both technicians are right.

4 0
3 years ago
A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p
saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, x, of an SHM is given by

x = A\cos(\omega t)

A is the amplitude and \omega is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or \frac{\pi}{4} radian.

From trigonometry, \sin A =\cos B if A and B are complementary.

At t = 0, x = 3.5

3.5 = A\cos(\omega \times0)

A =3.5

So

x = 3.5\cos(\omega t)

At t = 0.12, x = 1.5

1.5 = 3.5\cos(0.12\omega)

\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286

0.12\omega =\cos^{-1}0.4286

0.12\omega = 1.13

\omega = 9.4

The period, T, is related to \omega by

T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67

5 0
3 years ago
Never text and drive! Discuss with your parents the use of your cell phone when driving. Explain to them the one situation when
DanielleElmas [232]
One reason that it would be appropriate to talk on your cell phone, could be reporting an accident. For instance, a car crash, a sign in the road anything regarding safety of drivers.
7 0
3 years ago
what does it mean if the coefficient of friction is greater? for example,a coefficient of friction of 0.1 compared to 0.7
Sergeu [11.5K]
Well, 0.1 is actually less than 0.7, but I understand what you're asking.

The coefficient of friction describes the relationship between two surfaces
that are sliding by each other.  The higher the coefficient of friction is, the
'rougher' the meeting is, and the harder it is for one to slide over the other. 
A skate blade against ice has a very low coefficient of friction.  Sandpaper
against blue jeans has a high coefficient of friction.

A higher coefficient of friction means that when one thing is sliding over
the other one, friction robs more energy from the motion.  It's harder to
push one thing over the other one, and when you let go, the moving one
slows down and stops sooner.

Air resistance is actually an example of friction.  It prevents falling things
from falling as fast as they would if there were no air. The coefficient of
friction when something moves through air is pretty low.  If the same
object were trying to move through molasses or honey, the coefficient
of friction would be greater.

Friction robs energy, and turns it into heat.  So, especially in machinery with
moving parts, we want to make the coefficient of friction between the moving parts
as small as possible.  That's what the OIL in a car's engine is for.
3 0
3 years ago
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