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3 years ago

Which explanation of the solar system best fits the observations of the planets and how they orbit the sun?

Physics

2 answers:

AlekseyPX3 years ago

6 0

Answer: C. The sun and planets formed from a collapsing spinning cloud of gas and dust.

Explanation:

The entire solar system was formed from solar nebula. A nebula is a cloud of dust and gases. When this cloud spins and contracts, a protostar is formed. The temperature rises and nuclear fusion kicks in forming a star. The rest of the matter forms planets and other celestial bodies of the star system. Similarly, the Sun was formed in solar nebula from collapsing of spinning cloud of dust and gas.

NeTakaya3 years ago

4 0

You're right, Answer C

The dust and gas accumulate to form a solar nebula, which later on creates the star and the planets.

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A biconvex lens is formed by using a piece of plastic(n=1.70).

creativ13 [48]

Answer:

$f =17.15\ cm$

Explanation:

given,

refractive index of lens, n = 1.70

Radius of curvature of front surface. R₁ = 20 cm

Radius of curvature of the back surface, R₂ = 30 cm

focal length= ?

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

R₁ = +20 cm

R₂ = -30 cm

n = 1.70

$\dfrac{1}{f}=(1.70-1)(\dfrac{1}{20}-\dfrac{1}{-30})$

$\dfrac{1}{f}=0.70 \times 0.0833$

$f = \dfrac{1}{0.7 \times 0.0833}$

$f =17.15\ cm$

the focal length of the lens is equal to 17.15 cm

3 0

3 years ago

The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's moti

Art [367]

Answer:

Explanation:

on USAtestprep

7 0

2 years ago

The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an

Mrrafil [7]

Answer:

$v_1 = 3.5 \ m/s$

Explanation:

Given that :

mass of the SUV is = 2140 kg

moment of inertia about G , i.e $I_G$ = 875 kg.m²

We know from the conservation of angular momentum that:

$H_1= H_2$

$mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2$

$2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2$

$1637.1 v_1$ $= 3841.575 \omega_2$

$\omega_2 = \frac{1637.1 v_1}{3841.575}$

$\omega _2 = 0.4626 \ v_1$

From the conservation of energy as well;we have :

$T_2 +V_{2 \to 3} = T_3 \\ \\ \\ \frac{1}{2} I_A \omega_2^2 - mgh =0$

$[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0$

$706.93 \ v_1^2 - 8657.49 =0$

$706.93 \ v_1^2 = 8657.49$

$v_1^2 = \frac{8657.49}{706.93 }$

$v_1 ^2 = 12.25$

$v_1 = \sqrt{ 12.25$

$v_1 = 3.5 \ m/s$

6 0

3 years ago

A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet

Drupady [299]

Answer:

See answer

Explanation:

The area of the circular loop is given by:

$A = \pi r^2$

The magnetic flux is given by:

$\phi = \int \vec{B} \cdot d\vec{A}$

$d\vec{A}$ is parallel to $\vec{B}$ and $\vec{B}$ is constant in magnitude and direction therefore:

$\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2$

Part A)

initially the flux is $\phi =\pi B r^2$

after the interval $\Delta t= 2.4 [m/s]$

the flux is

$\phi = 0$

now, the EMF is defined as:

$\epsilon =- \frac{d \phi}{dt}$ ,

if we consider $\Delta t= 2.4 [m/s]$ very small then we can re-write it as:

$\epsilon =- \frac{\Delta \phi}{\Delta t}$

$\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12$

then:

$\epsilon =- \frac{-0.12}{0.0024} = 50 [V]$

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

3 0

3 years ago

Calculate the electric potential energy in a capacitor that stores 4.0 10-10 C of charge at 250.0 V

Arada [10]

Q = C.v
v = Q/C
v = 4 × 10^(-10)/250
= 4 × 10^(-10)/2.5 × 10^2
= 1.6 × 10^(-12) volt

7 0

3 years ago

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