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3 years ago

7

Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23

,500 km
from the center.
What is the ratio of the orbital period of Deimos to that of Phobos?

1 answer:

Sloan [31]3 years ago

8 0

Answer:

The ratio is $\frac{T_1}{T_2} = 3.965$

Explanation:

From the question we are told that

The radius of Phobos orbit is R_2 = 9380 km

The radius of Deimos orbit is $R_1 = 23500 \ km$

Generally from Kepler's third law

$T^2 = \frac{ 4 * \pi^2 * R^3}{G * M }$

Here M is the mass of Mars which is constant

G is the gravitational constant

So we see that $\frac{ 4 * \pi^2 }{G * M } = constant$

$T^2 = R^3 * constant$

=> $[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3$

Here $T_1$ is the period of Deimos

and $T_1$ is the period of Phobos

So

$[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}$

=> $\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]$

=> $\frac{T_1}{T_2} = 3.965$

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The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t

slamgirl [31]

Answer:

The charge is $2.75\times10^{-13}\ C$

Explanation:

Given that,

Distance = 2.5 mm

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We need to calculate the linear charge density

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$E=\dfrac{2k\lambda}{r}$

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$\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}$

$\lambda=1.1\times10^{-10}\ C/m$

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Using formula of charge

$Q=\lambda\timesL$

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$Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})$

$Q=2.75\times10^{-13}\ C$

Hence, The charge is $2.75\times10^{-13}\ C$

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3 years ago

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den301095 [7]

Answer:

68 m

Explanation:

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Initially, the turd was also flying horizontally with the birdman, so the initial velocity of the turd is the same as the horizontal velocity of the birdman, i.e In the horizontal direction, $u_0=17$ m/s.

In the vertical direction, u = 0,

The distance to be traveled, in the direction of application of force, is equals to the height of the turd, i.e

s= 78 m

Let t be the time taken to cover a distance of 78 m.

Now, applying the equation of motion in the vertical direction,

$s=ut+\frac 12 at^2$

where u is the initial velocity and a is the acceleration due to gravity in the direction of displacement,s.

Here, $a=g=9.81 m/s^2$ , so

$78=0\times t +\frac 12 (9.81)t^2$

$\Rightarrow t^2=(78\times2)/9.81$

$\Rightarrow t = 4$ seconds.

Hence, the time taken to reach the ground is 4 seconds.

As the horizontal speed, $u_0=17$ m/s, is constant throughout the journey, so

the horizontal distance covered by turd

$= u\times t$

$= 17 \times 4 = 68$ m.

So, the distance of landing from the start of the field is 68 m as the birdman releases a turd directly above the start of the field.

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3 years ago

Who is the founder of operant conditioning?

Gemiola [76]

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Hope this helps!

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3 years ago

This is physics 11th grade and a homework question I don’t understand how to do this or what the question is asking me

Alexxx [7]

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Frequency = 1.8 Hz

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Period = 0.056 s

b) When we differenctiate displacement with respect to time, the result is velocity.

Recall, period = 1/f = 5/9 cycles

1/4 cycle behind = 1/4 x 5/9 = 5/36

It is delayed with 5/36 sec with respect to displacement.

5/36 sec = 0.139 sec

Acceleration = first derivative of velocity = second derivative of displacement = 1/4 cycle behind velocity = 1/2 cycle behind displacement =

5/36 = 0.139 sec delayed with respect to velocity

= 5/18 = 0.2777 secs delayed with respect to displacement

Thus, the number of seconds out of phase with the displacements is 0.278 seconds

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T = 2π√length/local gravity). We would calculate the local gravity.

From the information given,

length = 0.2

T = P = 5/9

Thus,

5/9 = 2π√0.2/local gravity)

(5/9)/2π = √0.2/local gravity

Square both sides. It becomes

[(5/9)/2π]^2 = 0.2/local gravity

local gravity = 0.2/[(5/9)/2π]^2

local gravity = 25.56 m/s^2

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1 year ago

A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o

gulaghasi [49]

A. 0.77 A

Using the relationship:

$P=\frac{V^2}{R}$

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

$P_1=I^2 R_1$

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

$P_2=I^2 R_2$

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

$P=\frac{E}{t}$

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0

3 years ago