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3 years ago

Which of the Jovian planets have rings?

Physics

2 answers:

Sedbober [7]3 years ago

8 0

All four "Jovian planets" or "large planets" have rings on them and they are Saturn, Uranus, Jupiter and Neptune

stira [4]3 years ago

7 0

Jupiter , Saturn , Uranus , Neptune .

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PLEASE HURRYYYYYY:

Juliette [100K]

Answer:

Explanation:

4 0

4 years ago

Two billiard balls with the same mass undergo a perfectly elastic head-on collision. if one ball's initial speed was 2 m/s, and

My name is Ann [436]

<span>The ball with an initial velocity of 2 m/s rebounds at 3.6 m/s
The ball with an initial velocity of 3.6 m/s rebounds at 2 m/s

There are two principles involved here
Conservation of momentum and conservation of energy.
I'll use the following variables
a0, a1 = velocity of ball a (before and after collision)
b0, b1 = velocity of ball b (before and after collision)
m = mass of each ball.

For conservation of momentum, we can create this equation:
m*a0 + m*b0 = m*a1 + m*b1
divide both sides by m and we get:
a0 + b0 = a1 + b1

For conservation of energy, we can create this equation:
0.5m(a0)^2 + 0.5m(b0)^2 = 0.5m(a1)^2 + 0.5m(b1)^2
Once again, divide both sides by 0.5m to simplify
a0^2 + b0^2 = a1^2 + b1^2

Now let's get rid of a0 and b0 by assigned their initial values. a0 will be 2, and b0 will be -3.6 since it's moving in the opposite direction.
a0 + b0 = a1 + b1
2 - 3.6 = a1 + b1
-1.6 = a1 + b1
a1 + b1 = -1.6

a0^2 + b0^2 = a1^2 + b1^2
2^2 + -3.6^2 = a1^2 + b1^2
4 + 12.96 = a1^2 + b1^2
16.96 = a1^2 + b1^2
a1^2 + b1^2 = 16.96

The equation a1^2 + b1^2 = 16.96 describes a circle centered at the origin with a radius of sqrt(16.96). The equation a1 + b1 = -1.6 describes a line with slope -1 that intersects the circle at two points. Those points being (a1,b1) = (-3.6, 2) or (2, -3.6). This is not a surprise given the conservation of energy and momentum. We can't use the solution of (2, -3.6) since those were the initial values and that would imply the 2 billiard balls passing through each other which is physically impossible. So the correct solution is (-3.6, 2) which indicates that the ball going 2 m/s initially rebounds in the opposite direction at 3.6 m/s and the ball originally going 3.6 m/s rebounds in the opposite direction at 2 m/s.</span>

6 0

4 years ago

Two students both kick a soccer ball. Havanna's ball travels three times as fast as Mitzi's ball. How does the kinetic energy of

ladessa [460]

Answer:

the answer is c. Havanna's soccer ball has 9 times the kinetic energy, because it travels at 3 times the speed

Explanation:

4 0

3 years ago

What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +

Mumz [18]

Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, $\sigma_n = 28 MPa$

There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

Traction vector on n, $T_n = \sigma \b n$

$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

$\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa$

Since $\tau \neq 0$ , there are shear stresses acting on the surface.

3 0

3 years ago

One method of determining the location of the center of gravity of a person is to weigh the person as he/she lies on a board of

andrey2020 [161]

Answer:

x = 1.018 m

Explanation:

given,

height of man = 190 cm

= 1.9 m

scale reading on left = 450 N

scale reading on the right = 390 N

Let center of gravity of man be x distance from feet, feet is on right side.

For system to be in equilibrium moment about center should be equal to zero.

∑M = 0

now,

450(1.9 - x ) - 390 × x = 0

450(1.9 - x ) = 390 × x

855 - 450 x = 390 x

840 x = 855

$x = \dfrac{855}{840}$

x = 1.018 m

hence, point of center of gravity from feet is equal to x = 1.018 m

7 0

4 years ago