Which of the following sets of characteristics describe what we know about the outer planets? (2 points)Lowest daily average temperature, smaller in size and few or no moons.
Many moons, smaller in size and a ring system.
Rocky surface, closest to the sun and larger in size,
Gaseous composition, larger size and many moons.

Answers

Answer 1

Answer:

Gaseous composition, larger size and many moons describe about the outer planets.

What are outer planets?

Jupiter, Saturn, Uranus, and Neptune are the four outer planets. They are all gas giants consisting primarily of hydrogen and helium. Their interiors are liquid and contain thick gaseous outer layers. Numerous moons and planetary rings consisting of dust and other particles are present on every one of the outer planets.

To know more about outer planets refer to :

brainly.com/question/18392639

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Answer 2

Answer:

Answer: D

Explanation: Gaseous composition, larger size and many moons

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A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pulley, which is a disk of radius 9.00 cm , has friction in its axle.What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium? (Answer should be in N m)

Answers

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

Answer:here to earn points

Explanation:

insulator allows the electric current to pass through it true or false please anyone please please please

Answers

I say the answer is eating chicken. Eating chicken provides many nutrients such as, protein, diabetes, and somminilla.

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. the bicycle stops in 5.0 s. how far did the bicycle travel during the 5.0 seconds of braking?

Answers

Distance traveled by the bicycle during the 5 seconds of braking is 22m

Explanation:

initial angular velocity= 2 rev/s

final angular velocity= 0 rev/s

Angular displacement Ф= $((wi+wf)/(2) )$ t

Ф= $((0+2)/(2) )5=5$ rev

so the distance travelled= 5(2πr)

distance=5(2π*0.7)

distance=22m

The bicycle traveled about 22 m during the 5.0 seconds of braking

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Further explanation

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

$\texttt{ }$

Centripetal Force can be formulated as follows:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

$\texttt{ }$

Given:

radius of wheel = R = 0.70 m

initial angular speed = ω = 2.0 rev/s = 4π rad/s

final angular speed = ωo = 0 rad/s

time taken = t = 5.0 s

Asked:

distance covered = d = ?

Solution:

$d = \theta R$

$d = (\omega + \omega_o)(1)/(2)t R$

$d = ( 4 \pi + 0 ) (1)/(2)(5.0)( 0.70 )$

$d = 4\pi (1.75)$

$d = 7\pi \texttt{ m}$

$d \approx 22 \texttt{ m}$

$\texttt{ }$

Learn more

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

The mass of a string is 20 g and it has a length of 3.2 m. Assuming that the tension in the string is 2.5 N, what will be the wavelength of a travelling wave that is created by a sinusoidal excitation of this string with a frequency of 20 Hz. Provide the wavelength in units of m. Please note: You do not include the units in your answer. Just write in the number.

Answers

Answer:

The wavelength of the wave is 1 m

Explanation:

Given;

mass of the string, m = 20 g = 0.02 kg

length of the string, L = 3.2 m

tension on the string, T = 2.5 N

the frequency of the wave, f = 20 Hz

The velocity of the wave is given by;

$v = \sqrt(T)/(\mu) {}$

where;

μ is mass per unit length = 0.02 kg / 3.2 m

μ = 6.25 x 10⁻³ kg/m

$v = \sqrt{(T)/(\mu) } \n\nv = \sqrt{(2.5)/(6.25*10^(-3)) } \n\nv = 20 \ m/s$

The wavelength of the wave is given by;

λ = v / f

λ = (20 m/s )/ (20 Hz)

λ = 1 m

Therefore, the wavelength of the wave is 1 m

In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for F⃗ ? Let U=0 when x=0. Express your answer in terms of α and x.

Answers

Answer

$\Delta U= \alpha (x^3)/(3) \n$

Explanation:

given

$F = -\alpha x^2 i$

where $\alpha = 12 N/m^2$

now we know

$\int\limits^W_0 {} \, dW = \int\limits^a_b {F.} \, dxi$ ..................(i)

where dx is infinitesimal distance

$W = \int\limits^a_b {-\alpha x^2} \, dx \n$

for x = a and b = 0

after integration we get

$W = -\alpha (x^3)/(3)$

we know work done by conservative force will be equals to negative of potential energy

$W = -\Delta U$

so we get

$-\Delta U= -\alpha (x^3)/(3) \n\n\Delta U= \alpha (x^3)/(3) \n$

Which correctly describes how the energy of a wave on the electromagnetic spectrum depends on wavelength and frequency?A.
Energy decreases with decreasing wavelength and decreasing frequency.
B.
Energy increases with decreasing wavelength and increasing frequency.
C.
Energy increases with decreasing wavelength and decreasing frequency.
D.
Energy decreases with increasing wavelength and increasing frequency.

Answers

Answer:

B. Energy increases with decreasing wavelength and increasing frequency.

Explanation:

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