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steposvetlana [31]
3 years ago
8

Compare and contrast the general characteristics of a gas giant planet and one of the inner planets (such as Mercury, Earth, Ven

us or Mars)? Give two differences and one similarity.
Physics
1 answer:
yKpoI14uk [10]3 years ago
6 0
The gas planets usually have extremely high gravitational pulls, the surface isn't solid (since its a gas planet), and gas planets are larger than the inner planets. 
<span>Similarities- These planets all have moons and they both revolve around the sun (obviously).


Hope this helps.</span>
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Diego is playing basketball. While running at 7 km/h toward Gino, he passes the ball to Gino horizontally. The ball travels at 2
enot [183]

TheThe  relative velocity of Deigo will be 27km/h .

<h3>What is relative Velocity? </h3>

The relative velocity is the velocity of an object with respect to another observer. It is the time rate of relative position of one object with respect to another object .

Vab = Va-Vb

Where a is object and b is observer .

We have given here the velocity of ball 20km/ h

and the another velocity is 7km/ h

Let us assume that Diego is also moving in the direction where ball is moving so the relative velocity  will  be

 V = 20+7=27km/ h

When both observer and object is moving in same direction than there sum of velocity will be relative velocity ,if they are moving opposite than subtraction.

to learn more about Relative velocity click Here brainly.com/question/6070547

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3 0
1 year ago
The acceleration due to gravity on Jupiter is 2.5 times what it is here on earth. An object weighing 347.9 N here on earth will
inessss [21]

Answer:  weight on Jupiter = 869.75 N

              mass  on Earth = mass on Jupiter = 35.5 Kg

Explanation:

W = mg

W = weight

m = mass

g = gravitational acceleration [ on the Earth, g₁ = 9,8 N/kg ]

On the Earth,

G₁ = m x g₁  = 347,9 N

On the Jupiter,

G₂ = mg₂  

mass on the Earth = mass on the Jupiter !  

m = G₁ : g = 347.9 N : 9,8 N/kg = 35.5 kg

G2 : G1 = 2.5

G₂ = 2,5 G₁ = 2,5 x 347.9 N =  869,75 N

5 0
3 years ago
Two common ions of manganese are Mn2+ and Mn3+. Which binary ionic compound is likely to form? A) Mn2N3 B) MnBr4 C) MnCl D) MnO
meriva
Given that Oxygen has an oxidation state of  2 -, you can combine Mn 3+ with O 2- ions to form Mn2O3, and you can combine Mn 2+ with O 2- to form Mn2O2 which is MnO.

The other compounds imply oxidation states of N, Br and Cl that does not exist.

Therefore, the answer is the option D. MnO
4 0
3 years ago
Read 2 more answers
mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi
valina [46]

Answer:

The answer is

"x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs \frac{8}{3} feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\

The source of the hooks law is stable,

16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\

Number \frac{1}{2}  times the immediate speed, i.e .. Damping force

\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2}  \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\

The m^2+m+12=0 and m is an auxiliary equation,

m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \  m2 =\frac{-1 - \sqrt{47i}}{2}

Therefore, additional feature

x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]

Use the form of uncertain coefficients to find a particular solution.  

Assume that solution equation,

x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\

These values are replaced by equation ( 1):

\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\

Going to compare cos3 t and sin 3 t coefficients from both sides,  

The cost3 t is 3A + 3B= 20 coefficients  

The sin 3 t is 3B -3A = 0 coefficient  

The two equations solved:

3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\

Replace the very first equation with the meaning,

3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\

equation is

x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)

The ultimate plan for both the equation is therefore

x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.  

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\

x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t)  +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\

c_2=\frac{-64\sqrt{47}}{141}

x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))

5 0
3 years ago
If an object accelerates from rest, with a constant acceleration of 5.4 m/s2, what will its velocity be after 28s?
aleksklad [387]
Vf = Vi + at
Vf = 0 + 5.4•28
= 151.2m/s..
not sure if its right
6 0
3 years ago
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