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ArbitrLikvidat [17]
3 years ago
7

Two facts about Saturns largest moon

Physics
2 answers:
nikitadnepr [17]3 years ago
7 0

Titan is the largest moon of Saturn

Titan's diameter is 50 percent larger than that of Earth's moon. Titan is larger than the planet Mercury but is half the mass of the planet. Titan's mass is composed mainly of water in the form of ice and rocky material. Titan has no magnetic field.

Lena [83]3 years ago
6 0
The diameter of its sold body is 5,150 km(3,200 miles). The moon is called Ganymede.
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WILL GIVE BRAINLIEST! Briefly describe the three phases of self-regulation.<br> _PSYCHOLOGY_
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Phase 2. Performance control—This phase involves processes during learning and the active attempt to utilize specific strategies to help a student become more successful.

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6 0
3 years ago
Read 2 more answers
Someone can help me pls <br><br> Is physical science class
pentagon [3]

Answer:

<em> I can't see the picture</em>

Explanation:

6 0
3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
6. During an impact time casting 5 x 10-45 a gulf club exerts an average impar
Novay_Z [31]

Answer:

2.5 × 10-⁴¹ Ns

Explanation:

Impulse

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I = 2.5 × 10-⁴¹ Ns

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3 years ago
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tangare [24]

Answer:

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Explanation:

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