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3 years ago

13

Why does the ring on saturn spin

2 answers:

spayn [35]3 years ago

8 0

Answer: THERE IS NO GRAVITY IN SPACE SO ROCKS SPIN

Explanation:

Mnenie [13.5K]3 years ago

3 0

Answer:

The particles that make up Saturn's rings revolve around the planet. ... As Saturn spins around, so do its rings, rotating with the planet. More exactly, it is the millions of particles that make up these rings which rotate in the same direction as Saturn

Explanation:

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1. (1 points) What is the name of the drinking water supply well? a. VA1; b. VA24; c. VA19; d. VA40; e. VA18; 2. (1 points) What

Ksivusya [100]

Answer: b. 625.6

Explanation:

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3 years ago

Heat in the amount of 100 KJ is transferred directly from a hot reservoir at 1200 K to a cold reservoir at 600K.Calculate the en

Lina20 [59]

Answer:

0.0833 k J/k

Explanation:

Given data in question

total amount of heat transfedded (Q) = 100 KJ

hot reservoir temperature R(h) = 1200 K

cold reservoir temperature R(c) = 600 k

Solution

we will apply here change of entropy (Δs) formula

Δs = $\frac{Q}{R(h)}+\frac{Q}{R(c)}$

Δs = $\frac{-100}{1200}+\frac{100}{600)}$

Δs = $\frac{1}{12}$

Δs = 0.0833 K J/k

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3 years ago

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Two basic types of mechanical fuel injector systems?

REY [17]

Answer:

Single-point or throttle body injection. Port or multipoint fuel injection.

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3 years ago

Read 2 more answers

Write an application that solicits and inputs three integers from the user and then displays the sum, average, product, smallest

Ganezh [65]

Answer:

3423=6^H

Explanation:

6 0

3 years ago

A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio

natima [27]

Answer:

1) A=282.6 mm

2) $a_{max}=60.35\ m/s^2$

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm ------------1

V=2.5 m/s at x=225 mm ------------2

Where x measured from mid position.

We know that velocity in simple harmonic given as

$V=\omega \sqrt{A^2-x^2}$

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

$3.5=\omega \sqrt{A^2-0.15^2}$ ------3

$2.5=\omega \sqrt{A^2-0.225^2}$ --------4

Now by dividing equation 3 by 4

$\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}$

$1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}$

So A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

$3.5=\omega \sqrt{A^2-0.15^2}$

$3.5=\omega \sqrt{0.2826^2-0.15^2}$

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

$a_{max}=\omega ^2A$

$a_{max}=14.61 ^2\times 0.2826\ m/s^2$

$a_{max}=60.35\ m/s^2$

Time period T

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{14.609}$

T=0.42 sec

8 0

4 years ago