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IrinaK [193]
2 years ago
9

Which of the following statements about comets is true? A. Comets rarely fall into an orbit around the Sun. They usually enter t

he inner solar system only once and then are gone forever. B. Most comets have a circular orbit that keeps them in the space between Mars and Jupiter. C. Comets have very elliptical orbits that usually take them far beyond the orbit of Pluto, but also take them closer to the Sun than Earth. D. Comets have very elliptical orbits that usually take them closer to the Sun than Earth, but rarely do they get further away than Pluto.
Physics
2 answers:
Lelechka [254]2 years ago
7 0

the correct answer is C

GREYUIT [131]2 years ago
4 0
<h2>Right answer: Comets have very elliptical orbits that usually take them far beyond the orbit of Pluto, but also take them closer to the Sun than Earth</h2>

Comets are celestial bodies constituted by ice, dust and rocks that orbit around the Sun, after having been altered by the Oort cloud; following different trajectories that can be <u>highly eccentric elliptical</u><u> </u>(periodic trajectories), parabolic or hyperbolic.

One of the main characteristics of a comet is that it travels quite fast, on its way around the Sun and has a long tail. It should be noted that the tails of comets always go in the opposite direction to the Sun (due to the radiation pressure of sunlight).

Therefore, the correct option is C.

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Block A, mass 250 g , sits on top of block B, mass 2.0 kg . The coefficients of static and kinetic friction between blocks A and
masha68 [24]

Answer:

  F = 69.3 N

Explanation:

For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by

               fr = μ N

We define a reference system parallel to the floor

block B  ( lower)

Y axis  

            N - W₁-W₂ = 0

            N = W₂ + W₂

            N = (M + m) g

X axis

              F -fr = M a

for block A (upper)

X axis

              fr = m a                 (2)

so that the blocks do not slide, the acceleration in both must be the same.

Let's solve the system by adding the two equations

             F = (M + m) a          (3)

             a =\frac{F}{ M+m}

the friction force has the formula

            fr = μ N

             fr = μ (M + m) g

let's calculate

            fr = 0.34 (2.0 + 0.250) 9.8

            fr = 7.7 N

we substitute in equation 2

             fr = m a

             a = fr / m

             a = 7.7 / 0.250

             a = 30.8 m / s²

we substitute in equation 3

             F = (2.0 + 0.250) 30.8

             F = 69.3 N

5 0
2 years ago
A girl is shown at position A on a swing when the seat is directly below the support bar. The seat is then at height A as shown
MrRa [10]

Answer:

<u></u>

  • <u>1. The potential energy of the swing is the greatest at the position B.</u>

  • <u>2. As the swing moves from point B to point A, the kinetic energy is increasing.</u>

Explanation:

Even though the syntax of the text is not completely clear, likely because it accompanies a drawing that is not included, it results clear that the posittion A is where the seat is at the lowest position, and the position B is upper.

The gravitational <em>potential energy </em>is directly proportional to the height of the objects with respect to some reference altitude. Thus, when the seat is at the position A the swing has the smallest potential energy and when the seat is at the <em>position B the swing has the greatest potential energy.</em>

Regarding the forms of energy, as the swing moves from point B to point A, it is going downward, gaining kinetic energy (speed) at the expense of the potential energy (losing altitude). When the seat passes by the position A, the kinetic energy is maximum and the potential energy is miminum. Then the seat starts to gain altitude again, losing the kinetic energy and gaining potential energy, up to it gets to the other end,

7 0
2 years ago
Read 2 more answers
A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal
Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration (g=9.8 m/s^2) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

v=18.9 m/s

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

v_y^2 - u_y^2 = 2ah (1)

where

u_y = 16.4 m/s is the initial vertical velocity

v_y is the vertical velocity at height h

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, v_y = 0, so we can use the equation to find the maximum height

h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m

So, at half of the maximum height,

h = \frac{13.7}{2}=6.9 m

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s

And so, the speed of the stone at half of the maximum height is

v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

v_1 = 18.9 m/s

while the speed at half of the maximum height (part b) is

v_2 = 22.2 m/s

So the difference is

\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s

And so, in percentage,

\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0
2 years ago
According to the article, which pattern of brain waves are most conducive to studying new information?
sashaice [31]
Alpha brain waves are those most conducive to studying new information.

When consciously alert, we generally function along a beta brain rhythm. In diminishing this rhythm to alpha, we transition into a state of physical and mental relaxation that is ideal for learning new information and storing facts and  data. Studies have shown that the effect of decreasing brain rhythm is linked to feelings of increased mental clarity and remembrance. As it is a prime condition for synthetic thought and creativity, it becomes easier to visualize and create associations (information is better learned and absorbed by using such study methods). 

Hope this helps! :)
7 0
2 years ago
The 20-lb cabinet is subjected to the force F = (3 + 2t) lb, where t is in seconds. If the cabinet is initially moving down the
Andre45 [30]

Answer:

t₁ = 0.95 s

Explanation:

In this chaos we must use the definition of Newton's second law

      F = m a = m dv / dt

      dv = F dt / m

Let's replace and integrate, let's take the upward direction of the plane as positive, the force is positive

       dv = ∫ (3 + 2t) dt / m

       v = (3 t + 2 t²/ 2) /m

Let's evaluate between the lower limit t = 0 v = -6 ft / s (going down) to the upper limit   t = t and v = 0

       0 - (-6) = (3 (t- 0) + (t² -0)) / m

       t² + 3t -6m = 0

Let's look for the mass

      W = mg

      m = W / g

      m = 20/32

      m = 0.625 slug

Let's solve the second degree equation

     t² + 3t -3.75 = 0

     t = (-3 ± √ (32 + 4 1 3.75)) / 2

     t = (-3 ± 4,899) / 2

     t₁ = 0.95 s

     t₂ = -3.95 s

We take the positive time

6 0
2 years ago
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