3 sentences or more..........

Which of the following lies in the ecliptic plane?

babymother [125]

<h2>Answer: Earth's orbital path around the Sun</h2><h2></h2>

The <u>Ecliptic</u> refers to the orbit of the Earth around the Sun. Therefore, <u>for an observer on Earth it will be the apparent path of the Sun in the sky during the year, with respect to the "immobile background" of the other stars.</u>

<u />

It should be noted that the ecliptic plane (which is the same orbital plane of the Earth in its translation movement) is tilted with respect to the equator of the planet about $23\°$ approximately. This is due to the inclination of the Earth's axis.

Hence, the correct option is Earth's orbital path around the Sun.

7 0

4 years ago

What happens to the sum of the ball's kinetic energy and potential energy as the ball rolls from point A to point E? Assume ther

sesenic [268]

Answer:

C. The sum remains the same.

Explanation:

The sum of the kinetic and potential energy remains the same as the all rolls from point A to E.

We know this based on the law of conservation of energy that is in play within the system.

The law of conservation of energy states that "energy is neither created nor destroyed within a system but transformed from one form to another".

At the top of the potential energy is maximum
As the ball rolls down, the potential energy is converted to kinetic energy.
Potential energy is due to the position of a body
Kinetic energy is due to the the motion of the body

7 0

3 years ago

The electric potential difference between the ground and a cloud in a particular thunderstorm is 2.5 × 109 V. What is the magnit

muminat

Answer:

U = - 4 x $10^{-10}$

Explanation:

ΔV = potential difference = $2.5 * 10^9$ Volts

q = charge on electron = $-1.6 * 10^{-19}$ C

electric potential energy is given as

U = q ΔV = ( $-1.6 * 10^{-19}$ ) ( $2.5 * 10^9$ )

= - 4 x $10^{-10}$

6 0

3 years ago

As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horiz

Eddi Din [679]

a)E= U + K = $\frac{1}{2}$ kx² + $\frac{1}{2}$ mv²

The total energy of the system at any point in the motion is equal to the sum of the elastic potential energy of the spring, U, and of the kinetic energy of the mass, K:

E= U + K = $\frac{1}{2}$ kx² + $\frac{1}{2}$ mv²

where

'k' represents the spring constant

'x' is the compression/stretching of the spring with respect to its equilibrium position

'm' is the mass of the block attached to the spring

and 'v' is the speed of the block

b) <em>A=</em> $\sqrt{\frac{2E}{k}}$ <em> </em>

The amplitude of the motion compares to the most extreme displacement of the mass-spring system. The displacement of the system, x(t), at time t, for a simple harmonic oscillator is given by,

x= Asin(ωt+∅)

where

amplitude is 'A'

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency of the motion

t is the time

$\phi$ is the phase (we can take $\phi$ =0 )

The amplitude of the motion occurs when the displacement of the motion is maximum: x=A. Regarding energy, the mass-spring system is at its maximum displacement (x=A) when all the mechanical energy of the framework is elastic potential energy, so when the kinetic energy is zero:

K= $\frac{1}{2}mv^2$ =0

E= $\frac{1}{2}kA^2\\$ -->(1)

c) $v_{max}=\omega A$ <u></u>

When the elastic potential energy is zero, the maximum speed of the system occurs i.e U=0 and the kinetic energy is maximum, so:

U=0

E= $\frac{1}{2}mv_{max}^2$

According to the law of conservation of the mechanical energy, this energy must be equal to the energy of the system at its maximum displacement (1), so we can write

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2$

and solving for $v_{max}$ we find an expression for the maximum speed:

$v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{k}{m}}A=\omega A$

<h2><u></u> $v_{max}=\omega A$ <u></u></h2>

4 0

3 years ago

A 19.0-g bullet embeds itself in a 8.0-kg wooden block, which rests on a horizontal frictionless surface and is attached to a ho

Anton [14]

Answer:

d) 700 m/s

Explanation:

if k is the force constant and x is the maximum compression distance, then:

the potential energy the spring can acquire is given by:

U = 1/2×k×(x^2)

and, the kinetic energy system is given by:

K = 1/2×m×(v^2)

if Ki is the initial kinetic energy of the system, Ui is the initial kinetic energy of the system and Kf and Uf are final kinetic and potential energy respectively then, According to energy conservation:

initial energy = final energy

Ki +Ui = Kf +Uf

Ui = 0 J and Kf = 0J

Ki = Uf

1/2×m×(v^2) = 1/2×k×(x^2)

m×(v^2) = k×(x^2)

v^2 = k×(x^2)/m

= (500)×((21×10^-2)^2)/(19×10^-3 + 8)

= 2.75

v = 1.66 m/s

the v is the final velocity of the bullet block system, if m1 is the mass of bullet and M is the mass of the block and v1 is the initial velocity of the bullet while V is the initial velocity of the block, then by conservation linear momentum:

m1×v1 + M×V = v×(m1 + M) but V = 0 because the block is stationary, initially.

m1×v1 = v×(m1 + M)

v1 = v×(m + M)/(m1)

= (1.66)×(19×10^-3 + 8)/(19×10^-3)

= 699.86 m/s

≈ 700 m/s

Therefore, the velocity of the bullet just before it hits the block is 700 m/s.

5 0

3 years ago