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aksik [14]
3 years ago
10

Why are jovian planets so much larger than terrestrial planets?

Physics
1 answer:
Nataly [62]3 years ago
7 0

Explanation:

The temperature in the inner solar system was too high for light gases to condense, while in the outer solar system, the temperature was much lower, which allowed the Jovian planets to form, which grew enough to accumulate and retain the hydrogen gas that remained in the solar nebula, which led to its high levels of hydrogen and large size.

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A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find
Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

8 0
3 years ago
When all else remains the same, what effect would decreasing the focal length have on a convex lens?
aniked [119]
<h3>Answer;</h3>

<u>It would make the lens stronger. </u>

<h3>Explanation;</h3>
  • The focal length is the distance between the optical center or the center of the lens to the focal point of a convex or concave lens.
  • The power of the convex lens is lens ability to undertake refraction or bend light. It is given as the reciprocal of focal length.
  • Power of the lens = 1/ f; therefore the smaller the focal length the higher the power and the larger the focal length the lower the power.
  • Thus; decreasing the focal length of a convex lens makes the lens stronger.

4 0
3 years ago
Read 2 more answers
In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo
irinina [24]

Answer:

The torque about the origin is 2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}

Explanation:

Torque \overrightarrow{\tau} is the cross  product between force \overrightarrow{F} and vector position \overrightarrow{r} respect a fixed point (in our case the origin):

\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]

\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})

\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})

\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}

4 0
3 years ago
What causes winds to form?
nlexa [21]
Wind is caused by differences in the atmospheric pressure. When a difference in atmospheric pressure exists, air moves from the higher to the lower pressure area, resulting in winds of various speeds. On a rotating planet, air will also be deflected by the Coriolis effect, except exactly on the equator.
4 0
3 years ago
Years ago, a block of ice with a mass of about 20.0 kg was used daily in a home icebox. The temperature of the ice was 0.0°C whe
m_a_m_a [10]

Answer: The ice absorb 6671.1 kJ of thermal energy.

Explanation:

The conversions involved in this process are :

0.00^0C=273K

:H_2O(s)(273K)\rightarrow H_2O(l)(273K)

Now we have to calculate the enthalpy change.

\Delta H=n\times \Delta H_{fusion}

where,

\Delta H = enthalpy change = ?

m = mass of ice = 20.0 kg  = 20.0\times 10^3g    (1kg=1000g)

n = number of moles of ice= \frac{\text{Mass of ice}}{\text{Molar mass of water}}=\frac{20.0\times 10^3g}{18g/mole}=1.11\times 10^3mole

\Delta H_{fusion} = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

Now put all the given values in the above expression, we get

\Delta H=1.11\times 10^3mole\times 6010J/mole

\Delta H=6671100J=6671.1kJ     (1 kJ = 1000 J)

Therefore, the enthalpy change is,  6671.1 kJ

6 0
3 years ago
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