2 years ago

How can you convert Saturn radii to kilometers?

1 answer:

5 0

Multiply (Saturn radii) by (60,268) to get the distance in kilometers.

(This is the radius of the planet, not it's orbit.)

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Maru [420]

I have no clue sorry maybe try applying the answers and questions then I'll answer for u

7 0

3 years ago

A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.

Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder $(L)=5 m$

Foot the ladder is moving away with speed of $\frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s$

From diagram

$x^2+y^2=L^2$ ------1

at $x=3$

$y^2=25-9=16$

$y=4 m$

Now differentiating equation 1 w.r.t time

$2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0$

$x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}$

$3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}$

$\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s$

negative indicates distance is decreasing with time

5 0

2 years ago

in the system to the right, the pulleys are frictionless and the system hangs in equilibrium. Determine the values of each of th

irga5000 [103]

Look up pulleys problem through Khan academy and a video should pop up with a problem similar and you should be able to walk through it .

4 0

2 years ago

A boy drops a 0.10 kg stone down a 150 m well and listens for the echo. The air temperature is 20°C. How long after the stone is

S_A_V [24]

Use formula for Echo which is Velocity=2(Distance)/time so 343=2(150)/T 343T=300..T=300/343=0.9 seconds

4 0

2 years ago

The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would

natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a) $v_{es} =v_{rms}$ = 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

$v_{rms} = \sqrt{ \frac{3RT}{M}}$

So, $v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}$

multiply each side by M_{o2}, so we have

$v_{rms,o2}^2 *M_{o2} =3RT_{o2}$

solving for temperature T_{o2}

$T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}$

In the question given, $v_{rms} =v_{es}$

$T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}$

T_{o2} =1.58*10^5 K

7 0

3 years ago