All categories

3 years ago

Which of the following statements are true?Check all that apply.

Physics

1 answer:

MaRussiya [10]3 years ago

4 0

Answer:

1.The Sun is located at one of the foci of the planets' elliptical orbits.

2.The path of the planets around the Sun is elliptical in shape.

Explanation:

As per Kepler's law of planet motion we know that all planets revolve around the sun in elliptical path in such a way that position of Sun must be at one of the focii of the path

So all planets are in elliptical path always

Position of sun is always at one of the focus

so correct answer will be

1.The Sun is located at one of the foci of the planets' elliptical orbits.

2.The path of the planets around the Sun is elliptical in shape.

You might be interested in

At the end of cylindrical rod of length l = 1 m and mass M = 1 kg rotating horizontaly along the vertical axis in its center wit

matrenka [14]

Answer:

w = 0.943 rad / s

Explanation:

For this problem we can use the law of conservation of angular momentum

Starting point. With the mouse in the center

L₀ = I w₀

Where The moment of inertia (I) of a rod that rotates at one end is

I = 1/3 M L²

Final point. When the mouse is at the end of the rod

$L_{f}$ = I w + m L² w

As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved

L₀ = L_{f}

I w₀ = (I + m L²) w

w = I / I + m L²) w₀

We substitute the moment of inertia

w = 1/3 M L² / (1/3 M + m) L² w₀

w = 1 / 3M / (M / 3 + m) w₀

We substitute the values

w = 1/3 / (1/3 + 0.02) w₀

w = 0.943 w₀

To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s

w = 0.943 rad / s

3 0

3 years ago

A circuit containing an inductor and a capacitor in series is designed to have a resonant frequency of 4511 Hz. If the inductor

OLEGan [10]

Answer:

(e) $6.835\times 10^{-7}F$

Explanation:

At resonance we know that $X_l=X_C$

That is $\omega L=\frac{1}{\omega C}$

$\omega ^2=\frac{1}{LC}$

$\omega =\frac{1}{\sqrt{LC}}$

$f=\frac{1}{2\pi \sqrt{LC}}$

We have given resonance frequency f =4511 Hz and inductance L=1.82 mH

So $4511=\frac{1}{2\pi \sqrt{LC}}$

$LC=\frac{1}{4\pi ^2\times 4511^2}$

$LC=1.244\times 10^{-9}$

$C=\frac{1.244\times 10^{-9}}{1.82\times 10^{-3}}=0.6835\times 10^{-6}=6.835\times 10^{-7}F$

So option e is the correct answer

3 0

3 years ago

Guyz wqy-bdet-mby join m.e f.o.r f.u.n

sergey [27]

Answer:

What?

Explanation:

4 0

2 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

$-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

$\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}$

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

3 years ago

A 88.6-kg wrecking ball hangs from a uniform heavy-duty chain having a mass of 26.9kg . (Use 9.80m/s2 for the gravitational acce

Dafna11 [192]

Answer:

Tension maximum =1131.9 N

Tension minimum =868.28 N

Tension at 3/4= 1065.995 N

Explanation:

Given Mass of wrecking ball M1=88.6 Kg

Mass of the chain M2=26.9 Kg

Maximum Tension Tension max=(M1+M2) × (9.8 m/s²)

=(88.6+26.9) × (9.8 m/s²)

=115.5 × 9.8 m/s²

Tension maximum =1131.9 N

Minimum Tension Tension minimum=Mass of the wrecking ball only × 9.8 m/s²

=88.6 × 9.8 m/s²

Tension minimum =868.28 N

Tension at 3/4 from the bottom of the chain =In this part you have to use 75% of the chain so you have to take 3/4 of 26.9

= (3/4 × 26.9)+88.9) × 9.8 m/s²

= (20.175+88.6) × 9.8 m/s²

=(108.775) × 9.8 m/s²

=1065.995 N

6 0

3 years ago