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iVinArrow [24]
2 years ago
13

Why did astronomers suspect an eighth planet beyond uranus? how did they determine where to look for it? construct the correct e

xplanation.
drag the terms on the left to the appropriate blanks on the right to complete the sentences.

brightness - mass - orbit - position - kepler's third law - value - law of gravity


observation of uranus showed small but significant discrepancy between its predicted ______ and its actual ________ , which could mean that another object perturbed its motion.
in the mid-1800s, astronomers used the ________ to predict where a planet would have to be.
Physics
1 answer:
DerKrebs [107]2 years ago
7 0

Astronomers suspected an eighth planet beyond Uranus because of the irregularities in the motion of Uranus suggested that gravity from another planet was affecting it.

<h3>Who is an astronaut?</h3>

It should be noted that an astronaut simply means an individual who is trained to travel in spacecrafts.

In this case, astronomers suspected an eighth planet beyond Uranus because of the irregularities in the motion of Uranus suggested that gravity from another planet was affecting it.

In mid-1800s, astronomers used the orbits of most asteroids to predict where a planet would have to be.

Learn more about astronaut on:

brainly.com/question/2314879

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A moving particle fragments or decays into a particle moving at .53c, mass 135 MeV/c2, and a particle moving at .98c, mass 938 M
Svetllana [295]

Answer:

M = 1073 Mev/c2

u = 0.95 C        

Explanation:

given data:

m1 =135 Mev/c2

v1 = 0.53 c

m2 = 938 Mev/c2

v2 = 0.98 c

from conservation of momentum principle we have

\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = \frac{m1v1}{\sqrt{1-\frac{V1^2}{C^2}}} +\frac{m1v1}{\sqrt{1-\frac{V2^2}{C^2}}}

\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = \frac{135*0.53c}{0.848} +\frac{938*0.98c}{0.2}

\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = 4680.6 C   ...............1

Total mass of INITIAL particle M   =m1+m2 = 1073 Mev/c2

using equation 1

\frac{1073 u}{\sqrt{1-\frac{u^2}{C^2}}}= 4680.6C

solving for u we get

u = 0.95 C          

8 0
4 years ago
The figure shows an overhead view of a ring that can rotate about its center like a merry-go-round. Its outer radius R2 is 0.8 m
Shtirlitz [24]

The cat increase the kinetic energy of the cat-ring system is  42.4 J.

Mass of the merry-go-round = M = 7.6 kg

Outer radius of the merry-go-round = R2 = 0.9 m

Inner radius of the merry-go-round = R1 = R2/2 = 0.9/2 = 0.45 m

Moment of inertia of the merry-go-round = I

I = 3.8475 kg.m2

Mass of the cat = m = M/4 = 7.6/4 = 1.9 kg

Initially the cat is sitting at the outer edge that is at a distance of 'R2' from the center.

Initial moment of inertia of the system = I1

I1 = I + mR22

I1 = 3.8475 + (1.9)(0.9)2

I1 = 5.3865 kg.m2

Initial angular speed of the system =  title=View image!

1 = 7.6 rad/s

Now the cat walks to the inner edge of the merry-go-round therefore it is at a distance 'R1' from the center.

New moment of inertia of the system = I2

I2 = I + mR12

I2 = 3.8475 + (1.9)(0.45)2

I2 = 4.23225 kg.m2

New angular speed of the system =

I = I2

(5.3865)(7.6) = (4.23225)2

= 9.673 rad/s

Initial kinetic energy of the system = E1

E2 = 198 J

Amount by which the kinetic energy of the system increases =

E = E2- E1

title=View image!

E = 198 - 155.6

ΔE = 42.4 J

Amount by which the kinetic energy of the system increases when the cat crawls to the inner edge = 42.4 J.

Kinetic Energy:

Kinetic energy is the energy that an object possesses due to its motion. It is defined as the work required to accelerate an object of specified mass from rest to a specified velocity. After the body acquires this energy during acceleration, it retains this kinetic energy as long as the velocity does not change.

Learn about Kinetic Energy here:

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#SPJ4

3 0
1 year ago
=
pogonyaev

Answer:

c

Explanation:

4 0
2 years ago
Binding of a signaling molecule to which type of receptor leads directly to a change in the distribution of substances on opposi
Mrac [35]

When two sides of a membrane are in contact with each other, the distribution of ions will alter as a result of the binding of a signal molecule to a ligand-gated ion channel.

<h3>What is a ligand-gated ion channel?</h3>

Ligand-gated ion channels (LGICs) are membrane proteins that are structurally integral and feature a pore that permits the controlled passage of particular ions across the plasma membrane. The electrochemical gradient for the permeant ions drives the passive ion flux.

When a chemical ligand, such as a neurotransmitter, attaches to the protein, ligand-gated ion channels open. Changes in membrane potential cause voltage channels to open and close. When a receptor physically deforms, as in the case of pressure and touch receptors, mechanically-gated channels open.

Learn more about ligand-gated ion channel here:

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3 0
2 years ago
Four fixed point charges are at the corners of a square with sides of length L. Q1 is positive and at (OL) Q2 is positive and at
Ne4ueva [31]

Answer:

A) See Annex

B) Fq₁₂ = K *  Q₁*Q₂ /16 [N] (repulsion force)

C)  Fq₃₂  = K * Q₃*Q₂ /16 [N] (repulsion force)

D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)

E) Net force (its components)

Fnx = (2,59/64 )* K*Q²  [N] in direction of original Fq₃₂

Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂

Explanation:

For calculation of d (diagonal of the square, we apply Pythagoras Theorem)

d² = L² + L²    ⇒  d² = 2*L²     ⇒ d = √2*L²   ⇒ d= (√2 )*L

d = 4√2 units of length   (we will assume meters, to work with MKS system of units)

B) Force of Q₁ exerts on charge Q₂

Fq₁₂  = K * Q₁*Q₂ /(L)²     Fq₁₂ = K *  Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)

C) Force of Q₃ exerts on charge Q₂

Fq₃₂  = K * Q₃*Q₂ /(L)²     Fq₃₂  = K * Q₃*Q₂ /16  (repulsion force in the direction indicated in annex)

D) Force of -Q₄ exerts on charge Q₂

Fq₄₂ = K * Q₄*Q₂ / (d)²      Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)

E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)

Let´s take the force that  Q₄ exerts on Q₂  and Q₂ = Q  ( magnitude) and

Q₄ = -Q

Then the force is:

F₄₂ = K * Q*Q / 32       F₄₂  = K* Q²/32  [N]

We should get its components

F₄₂(x) = [K*Q²/32 ]* √2/2   and so is F₄₂(y)  =  [K*Q²/32 ]* √2/2

Note that this components have opposite direction than forces  Fq₁₂  and

Fq₃₂  respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively

In new conditions

Fq₁₂ = K *  Q₁*Q₂ /16    becomes  Fq₁₂ = K * Q²/ 16 [N]   and

Fq₃₂ = K* Q₃*Q₂ /16      becomes   Fq₃₂ = K* Q² /16  [N]

Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively

Then over x-axis we subtract Fq₃₂ - F₄₂(x)  = Fnx

and over y-axis, we subtract   Fq₁₂ - F₄₂(y) = Fny

And we get:

Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2  ⇒  Fnx =  K*Q² [1/16 - √2/64]

Fnx = (2,59/64 )* K*Q²

Fny has the same magnitude  then

Fny =(2,59/64 )* K*Q²

The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces

5 0
3 years ago
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