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3 years ago

The distance of Saturn from the sun is: < 1 A.U. > 1 A.U. = 1 A.U.

Physics

2 answers:

sertanlavr [38]3 years ago

6 0

Answer:

1 astronomical unit is the average distance from the Earth to the Sun; approximately 150 million km. At its closest point, Saturn is 9 AU, and then at its most distant point, it's 10.1 AU. Saturn's average distance from the Sun is 9.6 AU. We have written many articles about Saturn for Universe Today.

Explanation:

Hatshy [7]3 years ago

3 0

Please mark the first girl as brainest she tried her best

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In a circuit, energy is transferred to a charge.<br> Where is this energy transferred from?

vladimir1956 [14]

Answer:

Electric current.

Explanation:

The energy result from electric current resulting from potential differences between terminals which form an Electric circuit. This energy could come from different sources like chemical, wind, light

An electric circuit is one where there is movement of electrons;this electrons acquire charge which is energy. The electrons flow due to a potential difference; you have heard water flows from a higher position to a lower one freely. The highest height is said to be at higher potential and the lower point low potential.

So it's the same with electrons.

The formular for energy on charge is Q= I × t where I is electric current and t is time.

7 0

3 years ago

As an aid in working this problem, consult Interactive Solution 3.41. A soccer player kicks the ball toward a goal that is 20.0

alekssr [168]

Answer:

V=14.9 m/s

Explanation:

In order to solve this problem, we are going to use the formulas of parabolic motion.

The velocity X-component of the ball is given by:

$Vx=V*cos(\alpha)\\Vx=15.7*cos(31^o)=13.5m/s$

The motion on the X axis is a constant velocity motion so:

$t=\frac{d}{Vx}\\t=\frac{20.0}{13.5}=1.48s$

The whole trajectory of the ball takes 1.48 seconds

We know that:

$Vy=Voy+(a)*t\\Vy=15.7*sin(31^o)+(-9.8)*(1.48)=-6.42m/s$

Knowing the X and Y components of the velocity, we can calculate its magnitude by:

$V=\sqrt{Vx^2+Vy^2} \\V=\sqrt{(13.5)^2+(-6.42)^2}=14.9m/s$

6 0

4 years ago

To Mr. H's disgust, a 450-g black crow is raiding the recently-filled bird feeder. As Mr. H runs out the back door with his broo

bagirrra123 [75]

Answer:

Actually it's 2.50 m/s, sorry

Explanation:

It is solved by using momentum conservation equation

combined mass of crow and feeder = 450+670=1120 gm

let the recoil speed of feeder be v m/s

Then applying momentum conservation we get;

1120×1.5 = 670×v

v= 2.50 m/s

the speed at which the feeder initially recoils backwards = 2.50 m/s

5 0

3 years ago

What is the velocity of the object?

dmitriy555 [2]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2> $\huge\boxed{\text{V = 9.5 m/s}}$ </h2><h2>_____________________________________</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

${\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)$

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

$W\ =\ \frac{1}{2}\:6\:x\:30$

W = 90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

${\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2$

$\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}$

$\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

8 0

3 years ago

Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i

mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

$\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}$