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vekshin1
3 years ago
9

How does the location of the outer planets explain their composition and characteristics?

Physics
1 answer:
Elena-2011 [213]3 years ago
6 0

Jupiter, Saturn, Uranus and Neptune collectively make up the group known as the jovian planets. The general structures of the jovian planets are opposite those of the terrestrial planets. Rather than having thin atmospheres around relatively large rocky bodies, the jovian planets have relatively small, dense cores surrounded by massive layers of gas. Made almost entirely of hydrogen and helium, these planets do not have solid surfaces.

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Can someone help me find the which direction is north and which is south on this solenoid?
arsen [322]

Explanation:

Now, looking down the solenoid tube determine what direction is the winding. If clockwise in relation to the positive wire then is the south pole, if anti-clockwise then is the north pole. So, to summarize the magnetic south pole is always clockwise in relation to the positive wire.

7 0
3 years ago
The upper arm muscle is _______________ to the skin.
Nikolay [14]

Answer:

The pectoralis major, latissimus dorsi, deltoid, and rotator cuff muscles connect to the humerus and move the arm. The muscles that move the forearm are located along the humerus, which include the triceps brachii, biceps brachii, brachialis, and brachioradialis.

5 0
3 years ago
Read 2 more answers
If you want to double the kinetic energy of a gas molecule, by what factor must you increase its momentum?A) 16B) 2^(1/2)
worty [1.4K]

Answer: \sqrt{2}

The linear momentum p is given by the following equation:

p=m.v   (1)

Where m is the mass and v the velocity.

On the other hand, the kinetic energy K is given by:

K=\frac{p^{2}}{2m}   (2)

Which is the same as:

K=\frac{1}{2}m.v^{2}

Now, if we double the kinetic energy, equation (2) changes to:

2K=2\frac{p^{2}}{2m}  

2K=\frac{p^{2}}{m}   (3)

So, if we want to obtain the kinetic energy as shown in (3), the only option that works is increasing momentum by a factor of \sqrt{2} or 2^{1/2}:

Applying this in (2):

K=\frac{(\sqrt{2}p)^{2}}{2m}

K=\frac{(2p)^{2}}{2m}

K=\frac{p^{2}}{m}>>>As we can see, this equation is the same as equation (3)

Therefore, the correct answer is B

3 0
3 years ago
A block of mass M=10 kg is on a frictionless surface as shown in the photo attached. And it's attached to a wall by two springs
nordsb [41]

a.

  • i. the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s
  • ii. the angular velocity when the two springs are in parallel is 7.07 rad/s

b.

  • i. the speed of the block of mass when the springs are connected in series is 11.2 A m/s
  • ii. the angular velocity when the two springs are in series is 11.2 rad/s

<h3>a. </h3><h3>i. How to calculate the velocity of the mass when the springs are connected in parallel?</h3>

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k

= 2k

= 2 × 250 N/m

= 500 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where

  • k' = equivalent spring constant in parallel = 500 N/m,
  • A = maximum displacement of spring,
  • x = equilibrium position = 0 m,
  • M = mass of block = 10 kg and
  • v = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k'(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k'(A² - x²)/M]

v = √[500 N/m(A² - (0)²)/10]

v = √[50 N/m(A² - 0)]

v = [√50]A m/s

v = [5√2] A m/s

v = 7.07 A m/s

So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

<h3>ii. The angular velocity of mass when the springs are in parallel</h3>

Since velocity of spring v = ω√(A² - x²) where

  • ω = angular velocity of spring,
  • A = maximum displacement of spring and
  • x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 7.07 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 7.07 A m/s/√(A² - 0²)

ω = 7.07 A m/s/√(A² - 0)

ω = 7.07 A m/s/√A²

ω = 7.07 A m/s/A m

ω = 7.07 rad/s

So, the angular velocity when the two springs are in parallel is 7.07 rad/s

<h3>b. </h3><h3>i. How to calculate the velocity of the mass when the springs are connected in series?</h3>

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k

= 2/k

⇒ k" = k/2

k" = 250 N/m ÷ 2

= 125 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where

  • k" = equivalent spring constant in series = 125 N/m,
  • A = maximum displacement of spring,
  • x = equilibrium position = 0 m,
  • M = mass of block = 10 kg and
  • v' = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k"(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k"(A² - x²)/M]

v = √[125 N/m(A² - (0)²)/10]

v = √[125 N/m(A² - 0)]

v = [√125]A m/s

v = [5√5] A m/s

v = 11.2 A m/s

So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s

<h3>ii. The angular velocity of the mass when the springs are in series</h3>

Since velocity of spring v = ω√(A² - x²) where

  • ω = angular velocity of spring,
  • A = maximum displacement of spring and
  • x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 11.2 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 11.2 A m/s/√(A² - 0²)

ω = 11.2 A m/s/√(A² - 0)

ω = 11.2 A m/s/√A²

ω = 11.2 A m/s/A m

ω = 11.2 rad/s

So, the angular velocity when the two springs are in series is 11.2 rad/s

Learn more about speed of block of mass here:

brainly.com/question/21521118

#SPJ1

3 0
2 years ago
The molecule that functions as the reducing agent (electron donor) in a redox or oxidation-reduction reaction _____. Group of an
Elina [12.6K]

Answer:

b) loses electrons and loses potential energy

4 0
4 years ago
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