The large reservoir of comet nuclei far beyond Pluto, from which we believe new long-period comets come into the inner solar system, is called:

Answers

Answer 1

Answer:

Oort cloud

Explanation:

The large reservoir of comet nuclei far beyond Pluto, from which we believe new long-period comets come into the inner solar system, is called Oort cloud.

The comets hich have the large periods that means more than 200 years to orbit the Sun generaly comes from Oort cloud hich is also knon as the cometary cloud.


Related Questions

the masses of your hand and your notebook are quite small, so the force of attraction between them is

Answers

The force of attraction is decreased

please halp me solve this question!

Answers

current in 3ohm resistor is 0.9

Explanation:

total

Angie walked a distance of 90 meters east in 70 seconds. What was her
velocity?
A. 0.78 m/s east
B. 1.3 m/s east
O C. 7 m/s east
D. 9 m/s east

Answers

Answer:

1.3 m/s east

Step-by-step explanation:

90/70 = 1.3

Her velocity is 1.3

I let go of a piece of bread from a balcony. A bird flying 5.0 m overhead sees me drop it, and starts to dive straight down towards the bread the instant that I release it. She catches it after it falls 3.0 m. Assuming she accelerates constantly from rest (v0 = 0) at the time I let go of the bread, what is her acceleration? Show your work

Answers

This question can be solved using the equations of motion. There are two scenarios where the equations of motion can be used. The first scenario is the free-fall motion of the piece of bread. The second scenario is the uniformly accelerated motion of the bird.

The acceleration of the bird is  "a = 26.13 m/s²".

First, we will calculate the time taken by the bread to fall 3 m. Using the second equation of motion for this free-fall motion:

[tex]h = v_it + \frac{1}{2}gt^2[/tex]

where,

h = height fall = 3 m

vi = initial velocity = 0 m/s

g = acceleration due to gravity = 9.8 m/s²

t = time taken = ?

Therefore,

[tex]3\ m = (0\ m/s)t+\frac{1}{2}(9.8\ m/s^2)t^2\\t = \sqrt{\frac{(3\ m)(2)}{9.8\ m/s^2}}\\\\t = 0.78\ s[/tex]

The bird took the same time to catch the bread. Now applying the second equation of motion to the bird's motion:

[tex]s = v_it + \frac{1}{2}at^2[/tex]

where,

s = distance covered by the bird = 5 m + 3 m = 8 m

vi = initial velocity of the bird = 0 m/s

a = acceleration of the bird = ?

t = time taken = 0.78 s

Therefore, using these values we get:

[tex]8\ m = (0\ m/s)(0.78\ s)+\frac{1}{2}a(0.78\ s)^2\\\\a = \frac{16\ m}{(0.78\ s)^2}[/tex]

a = 26.13 m/s²

Learn more about the equations of motion here:

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Using a table for standard reduction potentials, calculate the minimum voltage needed to electrolyze potassium chloride into its elements (K(s) and Cl2(g)). Question 7 options: A) –1.57 V B) –4.29 V C) 1.57 V D) 4.29 V

Answers

Answer:  B  –4.29 V

Explanation:

I just took the quiz and I got a hundred percent make sure you pick B!!

Answer:

B  –4.29 V ;)

Explanation:

A force of 5N accelerates a mass by 2 m/s². What will be the acceleration if the force and mass were increased to twice their original value?

Answers

Answer:

4m/s4

Explanation:

Que. I : A mass of 10kg is suspended from the end of a steel of length 2m and radius 1mm, what is the elongation of the rod beyond its original length?

Que 2 : A pressure of sea water increases by 1.0atm for each 10metres increase in the depth. by what what percentage is the density of water increased in the deepest ocean of about 12km; compressibility = 5.0 × 10^-5 ​

Answers

Question 1; The elongation of the steel is approximately 0.3123 mm

Question 2; The percentage the density of water increased in the deepest

ocean is approximately 6.4%

The strategy of obtaining the above solution is presented as follows;

Que. 1; The given parameters are;

The mass of the suspended block, m = 10 kg

The length of the steel, l = 2 m

The radius of the steel, r = 1 mm = 1 × 10⁻³ m

The modulus of elasticity of steel, E = 200 GPa = 200 × 10⁹ Pa

The stress, σ, on the steel due to the mass, m, is given as follows;

[tex]\mathbf{\sigma = \dfrac{F}{A}}[/tex]

Where;

F = The force acting on the steel = The weight of the mass

A = The cross sectional area of the steel = π·r²

∴ F = 10 kg × 9.81 m/s² = 98.1 N

A = π × (1 × 10⁻³)² = 3.14159 × 10⁻⁶ m²

Therefore;

σ = 98.1 N/(3.14159 × 10⁻⁶ m²) ≈ 31,226,226.2 Pa

We have;

[tex]\mathbf{ E = \dfrac{\sigma}{\epsilon}}[/tex]

From which we have;

[tex]\epsilon = \dfrac{\sigma}{E}[/tex]

Where;

= The tensile strain = Δl/l

Δl = The elongation of the steel

Therefore;

∈ = 31,226,226.2/(200 × 10^9) = 0.00015613113

∴ Δl = 0.00015613113 × 2 m = 0.00031226226 m = 0.31226226 mm

The elongation of the steel, Δl = 0.31226226 mm ≈ 0.3123 mm

Question 2

The given parameters are;

The change in pressure per unit depth, Δp = 1.0 atm per 10 meters

The depth of the ocean = 12 km = 12,000 m

The compressibility = 5.0 × 10⁻⁵

The formula for compressibility, C, is presented as follows;

[tex]C = \dfrac{1}{V} \times \dfrac{\partial V}{\partial P}[/tex]

The change in pressure, [tex]\partial P[/tex] = 12,000 m × 1.0 atm/(10 m) = 1,200 atm

For a unit volume, V = 1 m³

We get;

[tex]5 \times 10^{-5} = \dfrac{1}{1} \times \dfrac{\partial V}{1,200}[/tex]

[tex]\partial V[/tex] = 5 × 10⁻⁵ m³/(atm) × 1,200 = 0.06 m³

The volume occupied 1 m³ at 12,000 km depth = V - [tex]\partial V[/tex]

∴ The volume occupied 1 m³ at 12,000 km depth = 1 m³ - 0.06 m³ = 0.94 m³

The percentage density increase, [tex]\partial[/tex]ρ% = (m/0.94 - m/1)/m/1 × 100

∴ (1/0.94 - 1/1)/1/1 × 100 ≈ 6.4%

The percentage increase in density  ≈ 6.4%

Learn more about elongation here;

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1. A bicycle initially moving with a velocity
5.0 m s-1 accelerates for 5 s at a rate of 2 m s? Wh
will be its final velocity ?

Answers

Answer:

[tex]\boxed {\boxed {\sf 15 \ m/s \ or \ 15 \ m*s^{-1}}}[/tex]

Explanation:

We are asked to find the final velocity. We are given the acceleration, time, and initial velocity, so we can use the following kinematics formula.

[tex]v_f= v_i+ at[/tex]

In this formula, [tex]v_f[/tex] is the final velocity, [tex]v_i[/tex] is the initial velocity, [tex]a[/tex] is the acceleration, and [tex]t[/tex] is the time.

The bicycle has an initial velocity of 5.0 m *s⁻¹ or m/s, acceleration of 2 m/s², and a time of 5 seconds.

[tex]\bullet \ v_i = 5.0 \ m/s \\\bullet \ a= 2\ m/s^2\\\bullet \ t= 5 \ s[/tex]

Substitute the values into the formula.

[tex]v_f=5.0 \ m/s + ( 2\ m/s^2 * 5 \ s)[/tex]

Solve inside the parentheses.

[tex]\frac {2 \ m}{s^2}* 5 \ s = \frac{ 2 \ m}{s} * 5 = \frac{ 10 \ m}{s} = 10 \ m/s[/tex]

[tex]v_f= 5.0 \ m/s + (10 \ m/s)[/tex]

Add.

[tex]v_f= 15 \ m/s[/tex]

The units can also be written as:

[tex]v_f= 15 \ m*s^{-1}[/tex]

The bicycle's final velocity is 15 meters per second.

What is the minimum value of force acting between two charges placed at 1 m apart from each other?
(a)Ke²
(b)Ke
(c)Ke/4
(d)Ke² /2

Answers

Answer:

Ke²

Explanation:

So,

q1 = e

q2 = e

r = 1m

By coulumb's law,

F = K (q1q2/r²)

F = K (e)(e)/(1)²  

F = Ke²  

 

Option(a)

Define Metrology
define Metrology ​

Answers

Answer:

the scientific study of measurement.

sort out electric current as fundamental or derived unit.​

Answers

Answer:

electric current is derived unit.

Explanation:

According to the definition of electric current, it appears to be a derived quantity. Charge on the other hand seems more fundamental than electric current.

The answer is electric current

Because it shows that the unit and it sorted out the electric

Lúc 7g bạn an đi từ nhà đến trường với tóc độ trung bình là 20km/h . Bạn đến trường lúc 7g20. Tính khoảng cách từ nhà tới trường?

Answers

Answer:

Distance = 6.667 kilometres

Explanation:

Given the following data;

Speed = 20 km/h

Departure time = 7:00

Arrival time = 7:20

Time taken = 20 minutes

To calculate the distance travelled from home to school;

First of all, we would have to convert the value of time in minutes to hours.

Conversion:

60 minutes = 1 hour

20 minutes = X hours

Cross-multiplying, we have;

X = 20/60 = 1/3 hours

Mathematically, the distance travelled by an object is calculated by using the formula;

Distance = speed * time

Distance = 20 * 1/3

Distance = 20/3 =

Distance = 6.667 kilometres

what are the examples of Inertia of motion​

Answers

Seat belts tighten in a car when it stops quickly.

Men in space find it more difficult to stop moving because of a lack of gravity acting against them.

When playing football, a player is tackled, and his head hits the ground.

EXAMPLES ARE GIVEN BELOW:

HOPE IT'S HELPFUL FOR U MATE

una caja en reposo se traslada 93 cm con un peso de 67N en un tiempo de 9,89h.¿cual es la aceleración la masa y la fuerza de dicho objeto

Answers

Answer:

a. Acceleration, a = 1.47 * 10^{-9} m/s²

b. Mass = 4.57 * 10^{10} kilograms

c. Force = 67.12 Newton

Explanation:

Given the following data;

Distance = 93 cm to meters = 93/100 = 0.93 meters

Weight = 67 N

Time = 9.89 hours to seconds = 35604 seconds

Initial velocity = 0 m/s (since it's starting from rest)

Acceleration due to gravity, g = 9.8 m/s

a. To find the acceleration, we would use the second equation of motion;

[tex] S = ut + \frac{1}{2} at^{2} [/tex]

Where;

S is the distance covered or displacement of an object.

u is the initial velocity.

a is the acceleration.

t is the time.

Substituting the values into the equation, we have;

[tex] 0.93 = 0*35604 + \frac{1}{2} * a*35604^{2} [/tex]

[tex] 0.93 = 0 + \frac{1}{2} * 1267644816a [/tex]

[tex] 0.93 = 633822408a [/tex]

[tex] Acceleration, a = \frac{0.93}{633822408} [/tex]

Acceleration, a = 1.47 * 10^{-9} m/s²

b. To find the mass

Weight = mass * acceleration due to gravity

67 = mass * 1.47 * 10^{-9}

[tex] Acceleration, a = \frac{67}{1.47 * 10^{-9}} [/tex]

Mass = 4.57 * 10^{10} kilograms

c. To find the force;

Force = mass * acceleration

Force = 4.57 * 10^{10} * 1.47 * 10^{-9}

Force = 67.12 Newton

If Earth's gravity pulls an object, causing it to accelerate to the ground, what

must be true about Earth?

A. It accelerates just as quickly in the direction away from the object.

B. It is being pulled toward the object by the object's gravity.

C. It accelerates just as quickly in the direction of the object.

D. It is being pushed away from the object by that same force.

Answers

Ans

It is being pulled toward the object by the objects gravity

Gravity:-

Sir Eizak Newton Founded the gravity.Gravity is a force between any object and earth by which they pull each other .The Acceleration due to gravity is represented by g =9.8m/s^2

A piece of gum becomes stuck upon a skateboard's wheel. What is the centripetal acceleration of the piece of gum if the wheel's radius is 30 mm and the tangential velocity is 0.5 m/s?

Answers:
a. 8.33 m/s^2
b. 0.5 m/s^2
c. 30 mm/s^2
d. 0.33 m/s^2
e. 83.20 m/ft^2

Answers

Answer:

a

Explanation:

a_c = v_t^2/r

a_c = (0.5)^2/0.03

a_c = 8.33 m/s^2

Bonds Quick Check
Metallic bonds are responsible for many properties of metals, such as conductivity. Why is this possible? (1 point)
A Metals have low electronegativity, so they are conductive because they pass charges easily.

Two metals bonded together are going to be more conductive than a metal bonded with a
nonmetal
C An attraction between a positive charge and a negative charge is created.
The bonds can shift because valence electrons are held loosely and move freely.

Answers

Answer: The bonds can shift because valence electrons are held loosely and move freely.

Explanation:

Which scientist proposed the first atomic theory?

Answers

Answer:

The Greek philosophers leucippus and Democritus

Answer:

Democratus

Explanation:

I hope it help you

A comet of mass 2 × 10^8 kg is pulled toward the star. If the comet's initial velocity is very small, and the comet starts moving toward the star from 700,000,000 km away, how fast is it going right before it hits the surface of the star? (Assume that it does not lose any mass by melting as it approaches the star.)

Answers

Answer:

The speed of the comet at the surface of the star is approximately 1,208,694.7 m/s

Explanation:

Question parameter obtained online; The mass of the star, M = 5 × 10³¹ kg

Explanation;

The given mass of the comet, m = 2 × 10⁸ kg

The initial velocity of the comet, v → 0

The distance of the comet from the star, d = 700,000,000 km

The gravitational potential at d = G·M·m/d

The kinetic energy of the comet, K.E. = m·v²/2

The kinetic energy of the comet at d = m·(0)²/2 = 0

The gravitational potential at the surface of the star, R = G·M·m/R

The kinetic energy of the comet at the surface of the star, R = m·(v)²/2 = 0

Where;

M = The mass of the star = 5 × 10³¹ kg

[tex]M_{Sun}[/tex] = The mass of the Sun = 1.989 × 10³⁰ kg

M/[tex]M_{Sun}[/tex] = 5 × 10³¹/(1.989 × 10³⁰) ≈ 25

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

R = The radius of the star

Therefore, we have;

m·(0)²/2 - G·M·m/d = m·v²/2 - G·M·m/R

∴ v = √((G·M·m/R - G·M·m/d)×2/m) = √(2·G·M(1/R - 1/d))

Therefore; v = (2 × 6.67430 × 10⁻¹¹ × 5 × 10³¹ × (1/R - 1/700,000,000,000))

v = 81696389149.1×√(1/R - 1/700,000,000,000).

The speed of the comet at the surface of the star, v = 81696389149.1×√(1/R - 1/700,000,000,000)

The mass radius relationship is given as follows;

[tex]\dfrac{R}{R_{Sun}} = 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]

[tex]R = R_{Sun} \times 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]

The radius of the Sun = 696,340,000 M

∴ R ≈ 696,340,000 × 1.3 × √(25.14) = 4538865694.76

R = 4538865694.76 m

v = 81696389149.1×√(1/4538865694.76 - 1/700,000,000,000) ≈ 1208694.7  m/s

determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3​

Answers

Answer:

divide the density of solution by density of water

EXPLANATION:

LIKE:

1.25÷1000kgm-3

A lady walks 10 m to the north, then she turns and continues walking 30 m due east.

Determine her(a) distance covered

(b) displacement.

Answers

Answer:

The distance covered is 40 m and the displacement is 31,6m.

Explanation:

The distance covered is the sum of the two distances (10+30). The displacement is equal to the distance of the hipotenusa of the triangle that the two distances (10 m to north and 30m to east) create. Using the Pythagoras theorem the displacent is equal to the Square root of (30^2 +10^2) .

please answer
during a journey, a car travels at 40 km in 2.5 hours, next 62 km in 3 hours, then took a break for 30 minutes, again travelled the last 120 km in 3.2 hours. calculate the average speed of the car during the journey.

Answers

average speed of the car is 23.9 km/h

Why does the value of g is move at polar region than at equator

Answers

Explanation:

the centrifugal force of the Earth's rotation forces the Earth's belly at the equator to grow even further. that extra material/mass is pulled in from the poles.

so, from equator to equator there is more mass and more gravity than from pole to pole.

what do you mean by 5meter length​

Answers

Answer:

h, nzk SC j AZ hsbzkzjx NM s in xbxnjxjx
it means that it is 5times greater than the platinim irridium rod kept in international bearuo of france

Sort the processes based on the type of energy transfer they involve. condensation freezing deposition sublimation evaporation melting thermal energy added thermal energy removed

Answers

Answer:

condensation - thermal energy removed

freezing -thermal energy removed

deposition - thermal energy removed

sublimation - thermal energy added

evaporation - thermal energy added

melting - thermal energy added

Explanation:

Thermal energy is heat energy. Processes in which heat is added involve the addition of thermal energy while processes in which heat energy is removed involves removal of thermal energy.

Condensation involves a change from gas to liquid, freezing involves a change from liquid to solid while deposition involves the settling of mobile particles at a place. All these processes involve a decrease in energy of particles.

On the other hand, sublimation is a direct change from solid to gas, melting involves a change from solid to liquid while evaporation involves a change from liquid to gas. All these processes occur when energy is added to the particles in a system.

Answer:

condensation - thermal energy removed

freezing -thermal energy removed

deposition - thermal energy removed

sublimation - thermal energy added

evaporation - thermal energy added

melting - thermal energy added

s27253129 ,,, message me please, I can't ask you my homework question in the comments :c

Answers

What’s the problem what do u need help on

La resistencia de un termómetro de platino es de 6Ω a30°C. Hallar su valor correspondiente a 100°C,sabiendo que el coeficiente de temperatura de resistividad del platino vale 0,00392°C^(-1).

Answers

Respuesta:

7,6 Ω

Explicación:

Paso 1: Información dada

Resistencia a 30 °C (R₀): 6 ΩCoeficiente de temperatura (α): 0,00392 °C⁻¹

Paso 2: Hallar la resistencia (R) a 100 °C

Podemos ver la relación entre la resistencia de un material y la temperatura usando la siguiente ecuación.

R = R₀ (1  + α × ΔT)

R = 6 Ω (1  + 0,00392 °C⁻¹ × (100 °C - 30 °C)) = 7,6 Ω

diffusion in physics​

Answers

Answer:

sorry but which class your talking 'bout

a small object is placed between two plane mirrors inclined at an angle of 60° to each other in a dark room how many images are seen explain​

Answers

Answer:

nothing

Explanation:

bocouse of darkness

Describe the responses of the human ear to sound waves coming from the
right side, left side, or in phase.

Answers

If the sound comes from the right side, the waves reach the right ear before the left ear. if the sound comes from the left side, the waves reach the left ear before the right ear. The difference between the phases of waves reaching both ears is detected by the ears and then interpreted by the brain

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What is the GCF of 1683t, 4085, and 68t??O 4O 483tO 8O 8837 the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is