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2 years ago

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Between 1992 and today, astronomers using large telescopes have discovered many icy pieces that orbit in the same region as the

orbit of Pluto. These are believed to be members of the

1 answer:

pantera1 [17]2 years ago

6 0

Icy pieces that orbit in the same region as the orbit of Pluto are believed to be members of the Kuiper belt.

<h3>What is the Kuiper belt?</h3>

The Kuiper belt is a shape-disc formation in the solar system observed from Neptune and more away from our sun.

The Kuiper belt consists of different celestial bodies including among others icy pieces, planets, and comets.

In conclusion, icy pieces that orbit in the same region as the orbit of Pluto are believed to be members of the Kuiper belt.

Learn more about the Kuiper belt here:

brainly.com/question/3955749

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What is the de Broglie wavelength for a proton with energy 50 keV? Due to the limitations of Canvas, please give the wavelength

ad-work [718]

Answer:

1.2826 x 10^-13 m

Explanation:

$\lambda = \frac{h}{\sqrt{2 m K}}$

Here, k be the kinetic energy and m be the mass

K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

m = 1.67 x 10^-27 kg

$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}$

λ = 1.2826 x 10^-13 m

6 0

3 years ago

Which is NOT true of the Big Bang Theory?

grigory [225]

Answer:

1. it explains what's happening im the universe now

3 0

3 years ago

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be $v_{AT}=v_A-v_T$ (<em>the train must see the car advancing at a lower speed</em>), where $v_A$ is the speed of the automobile and $v_T$ the speed of the train.

So we have $v_{AT}=(95km/h)-(75Km/h)=20Km/h$ .

So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s$

And in that time the car would have traveled (<em>relative to the ground</em>):

$d=v_At=(95Km/h)(0.065h)=6.175Km$

If they are traveling in opposite directions, <u>we have to do all the same</u> but using $v_{AT}=v_A+v_T$ (<em>the train must see the car advancing at a faster speed</em>), so repeating the process:

$v_{AT}=(95km/h)+(75Km/h)=170Km/h$

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s$

$d=v_At=(95Km/h)(0.00765h)=0.73Km$

5 0

3 years ago

. While a person lifts a book of mass 2 kg from the floor to a tabletop, 1.5 m above the floor, how much work does the the perso

viktelen [127]

Answer:

a) 29.4 J

b) - 29.4 J

Explanation:

Given:

Mass of the book, m = 2 kg

Height above the floor, h = 1.5 m

Now,

the work done by the person will be = Force applied on the book × displacement of the book

thus,

Work done by the person = mg × h

where, g is the acceleration due to gravity

thus, on substituting the values, we get

Work done by the person = 2 × 9.8 × 1.5 = 29.4 J

now,

for the force applied by the gravitational pull (downwards) the displacement is in opposite direction (upwards) to the force of the gravity.

Thus,

work done by the gravity will be negative

therefore, the work done by the gravity = - mg × h

or

work done by the gravity = - 29.4 J

5 0

3 years ago

Read 2 more answers

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s in 2.99 s.(a) fi

elena55 [62]

The angular acceleration of a rotating object is given by
$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$
where
$\omega_f$ is the final angular speed of the object
$\omega_i$ is its initial angular speed
$\Delta t$ is the time taken to accelerate

For the wheel in our problem, $\omega_f=11.1 rad/s$ , $\omega_i = 0$ and $\Delta t=2.99 s$ , so its angular acceleration is
$\alpha= \frac{11.1 rad/s-0}{2.99 s}=3.71 rad/s^2$

8 0

3 years ago