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Ivanshal [37]
3 years ago
6

____, one of Saturn's icy moons, is unusual in the solar system in that it has volcanic activity that ejects plumes of icy parti

cles into space.
Physics
1 answer:
blondinia [14]3 years ago
4 0

Answer: Enceladus

Explanation:

Enceladus is a small, icy body with an undergound ocean beneath its crust. Cassini discovered that geyser-like jets spew water vapor and ice particles. It is also the sixth largest moon in Saturn and just about a tenth of the largest moon in Saturn; Titan. It is often regarded as one of the most reflective body in the solar system as a result of its icy surface.

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Match the terms to the correct descriptions. Question 1 options:
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1) Refraction
2)Reflection
3)Concave
4)Convex

I took the test and got this right so you can believe me :)
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3 years ago
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How many joules of heat must be transferred to a 410-g aluminum pizza pan to raise its temperature from 32oC to 232oC? The speci
xxTIMURxx [149]

Answer:

recall that heat absorbed released is given by

Q = mc*(T2 - T1)

where

m = mass (in g)

c = specific heat capacity (in J/g-k)

T = temperature (in C or K)

*note: Q is (+) when heat is absorbed and (-) when heat is released.

substituting,

Q = (480)*(0.97)*(234 - 22)

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3 years ago
According to the rayleigh criterion, what is the shortest object we could resolve at the 25.0 cm near point with light of wavele
Blizzard [7]

Answer:

6.71 *10^{-5} rad

Explanation:

∅ = \frac{1.22*wavelength}{D} = \frac{1.22*550 * 10^{-9} }{25 * 10^{-2} }

∅ = 6.71 *10^{-5} rad

The minimum resolvable angle = 6.71 *10^{-5} rad

7 0
2 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

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3 years ago
consider the mirror from the last question. an object 4cm tall stands 10cm in front of a converging mirror of focal length 5cm.
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