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3 years ago

Which characteristic of the gas giants decreases with increasing distance from the sun?

Physics

2 answers:

hodyreva [135]3 years ago

7 0

Equator diameter and Number of moons. So both a and c would be the answer :)

anyanavicka [17]3 years ago

4 0

Gas giant are plants with mass 10 times bigger than the mass of the Earth. They are also called outer planets. There are four gas giant planets in our solar system: Jupiter, Saturn, Uranus and Neptune. The farther away from the Sun they are (the distance from the Sun is increased) the longer they orbit the Sun because of the great distance.

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water is a molecule of oxygen and hydrogen. The bonds that hold water molecules together are due to shared what

Olenka [21]

The bonds that hold water molecules together are due to shared electrons. Hope I helped!

8 0

2 years ago

Breanna is standing beside a merry-go-round pushing 19° from the tangential direction and is able to accelerate the ride and her

leva [86]

To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.

Torque in a body is defined as,

$\tau_l = F*d$

And in angular movement like

$\tau_a = I*\alpha$

Where,

F= Force

d= Distance

I = Inertia

$\alpha =$ Acceleration Angular

PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

$\tau= F*cos(19)*d$

On the other hand we have the speed data expressed in RPM, as well

$\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 1.0471rad/s$

Acceleration can be calculated by

$\alpha = \frac{\omega_f}{t}$

$\alpha = \frac{1.0471}{9}$

$\alpha = 0.11rad/s^2$

In the case of Inertia we know that it is equivalent to

$I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2$

$I = 1983.75kg.m^2$

Matching the two types of torque we have to,

$\tau_l=\tau_a$

$Fd=I\alpha$

$Fcos(19)*2.3=1983.75(0.11)$

$F=100.34N$

PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,

$W = \frac{1}{2}I\omega_f^2$

$W= \frac{1}{2}(1983.75)(1.0471)^2$

$W=1087.51J$

7 0

2 years ago

The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor re

mihalych1998 [28]

Answer:

Explanation:

Given

$\omega =13-\frac{1}{2}\cdot t^2$

Motor reverse its direction when \omega =0

$13-0.5t^2=0$

$26=t^2$

$t=\sqrt{26}=5.099\approx 5.1 s$

(b)

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega$

$\int d\theta =\int_{0}^{5.1}\omega dt$

$\int d\theta =\int_{0}^{t}(13-.05t^2)dt$

$\theta =(13t-0.1667\times t^3)_0^{5.1}$

$\theta =44.192^{\circ}$

4 0

2 years ago

What is the frequency of sound waves whose wavelength is 0.25 m on a day when the air temperature is 25 degree Celsius? Group of

Dvinal [7]

Answer:1384 Hz

Explanation:

Given

wavelength $(\lambda )$ =0.25 m

Temperature T $=25^{\circ}$

at $T=25^{\circ}$ velocity of sound is 346 m/s

and we know

$velocity=frequency\times \lambda$

$v=f\times \lambda$

$346=f\times 0.25$

f=1384 Hz

5 0

2 years ago

Read 2 more answers

a fan acquires a speed of 180 rpm in 4s, starting from rest. calculate the speed of the fan at the end of the 5th second startin

KengaRu [80]

Answer:

225 rpm

Explanation:

The angular acceleration of the fan is given by:

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_f$ is the final angular speed

$\omega_i$ is the initial angular speed

$\Delta t$ is the time interval

For the fan in this problem,

$\omega_i = 0\\\omega_f = 180 rpm\\\Delta t=4 s$

Substituting,

$\alpha = \frac{180-0}{4}=45 rpm/s$

Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:

$\omega' = \omega_i + \alpha t$

where

$\omega_i = 0\\\alpha = 45 rpm/s\\t = 5 s$

Substituting,

$\omega' = 0 + (45)(5)=225 rpm$

7 0

2 years ago