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2 years ago

Which of the following is not a characteristic of Jovian planets?

Physics

2 answers:

Katyanochek1 [597]2 years ago

7 0

The answer is C) thin atmospheres.

Hope it helps!

anyanavicka [17]2 years ago

3 0

I'll answer since there are no answers to refer to. A jovian planet is pretty much a planet that doesn't have a solid surface. It's a planet made up primarily of gases hydrogen, nitrogen, helium, methane and so on

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What is the average speed of a car that traveled 222.0 miles in 4.596 hours?

vivado [14]

48.3 mph
If you need more answers like this I suggest countcalculate .com

4 0

2 years ago

Please Help!!

kotegsom [21]

Answer:

Potential energy of spring = 24 Joules.

Explanation:

Given the following data;

Spring constant = 85N/m

Extension, e = 0.75m

Mass = 25kg

To find the potential energy of a spring

Potential energy of a spring is given by the formula;

P.E = ½ke²

Substituting into the equation, we have

P.E = ½*85*0.75²

P.E = 42.5 * 0.5625

P.E = 23.91 ≈ 24 Joules

P.E = 24 Joules

8 0

2 years ago

Seja uma carga Q=1.2×10(-8)C no vácuo, distando 40 cm de um ponto M e 50 cm de um ponto N . Determine os pontencias no ponto M e

Len [333]

<span>Americano, por favor?</span>

4 0

3 years ago

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang

koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × $10^{-4}$ Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux is express as

flux Ф = $\int \vec{B}.d\vec{A}$ .................1

and Ф = BA cos∅ ............2

so B = $\frac{\phi }{Acos\theta }$

and we know

A = ab

B = $\frac{\phi }{abcos\theta }$ ..............3

put here value

B = $\frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}$

solve we get

B = 0.692 T

8 0

2 years ago

A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show

Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

$x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]$

And the speed can be calculated as follows:

$v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]$

7 0

2 years ago