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irinina [24]
3 years ago
6

Look at the densities of the jovian planets given in figure 1. which of the following statements best describes the pattern of j

ovian planet densities?
Physics
2 answers:
barxatty [35]3 years ago
4 0

more than 5.5 times the density of water

is the answer.

avanturin [10]3 years ago
3 0
Given the densities of the jovian planets, the densities is just like random numbers, so the densities of the jovian planets have no concrete pattern due to its randomness. pattern can be either linear or geometric, where linear you see a common difference among data, while geometric you observed a common ratio
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Once again we have a skier on an inclined plane. The skier has mass M and starts from rest. Her speed at the bottom of the slope
mars1129 [50]

Answer:

v = 31.3 m / s

Explanation:

The law of the conservation of stable energy that if there are no frictional forces mechanical energy is conserved throughout the point.

Let's look for mechanical energy at two points, the highest where the body is at rest and the lowest where at the bottom of the plane

Highest point

       Em₀ = U = m g y

Lowest point

     Em_{f} = K = ½ m v²

As there is no friction, mechanical energy is conserved

       Em₀ = Em_{f}

       m g y = ½ m v²

       v = √ 2 g y

Where we can use trigonometry to find and

       sin 30 = y / L

       y = L sin 30

Let's replace

      v = RA (2 g L sin 30)

Let's calculate

      v = RA (2 9.8 100.0 sin30)

      v = 31.3 m / s

4 0
3 years ago
Please help me with this
Firlakuza [10]

Answer:

can't see anything sorry can't help

7 0
3 years ago
Read 2 more answers
In which situation is no work being done? A. a person carrying a box from one place to another B. a person picking up a box from
Anastaziya [24]

Picking up a box, pushing a box along the ground, and pulling a box along the ground (B, C, and D) definitely involve work being done.

If a person CARRIES a box from one place to another, AND keeps it at the same height during the entire carry, then no work is done. ( A )

6 0
4 years ago
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Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length firs
mafiozo [28]

Answer:

Magnitude of the vector is 23.75\ \text{units} and the direction is 35.23^{\circ}

Explanation:

Magnitude of first vector = |A| = 13\ \text{units}

Angle = \theta_1=27^{\circ}

Magnitude of second vector = |B| = 11\ \text{units}

Angle = \theta_2=45^{\circ}

x component of first vector

A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}

y component of first vector

A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}

x component of second vector

B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}

y component of first vector

B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}

Adding the magnitudes

C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}

C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}

Magnitude of the sum of the vectors would be

|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}

The direction would be

\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}

The magnitude of the vector is 23.75\ \text{units} and the direction is 35.23^{\circ}

4 0
3 years ago
How are you guys today? I'm just wasting points' to talk.
scoundrel [369]
I’m honestly bored & tiredddd
5 0
3 years ago
Read 2 more answers
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