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Zanzabum
3 years ago
15

The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the

planet, along a line connecting the centers of planet and moon, the net force on an object will be zero. Natividad, Joshua - [email protected] @theexpertta - tracking id: 1F76-1A-38-41-94BC-19489. In accordance with Expert TA's Terms of Service. copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account. show answer No Attempt 50% Part (a) Derive an expression for x.

Physics
1 answer:
seraphim [82]3 years ago
6 0

Answer:

x=\dfrac{D}{1+\sqrt{\dfrac{m}{M}}}

Explanation:

Mass of moon = m

Mass of planet =M

We know that gravitational force given as

F=G\dfrac{m_1m_2}{d^2}

F'=G\dfrac{m'M}{x^2}

F=G\dfrac{m'm}{(D-x)^2}

Given that force is zero so

F=F'

G\dfrac{m'm}{(D-x)^2}=G\dfrac{m'M}{x^2}

\dfrac{m}{(D-x)^2}=\dfrac{M}{x^2}

\dfrac{x}{D-x}=\sqrt{\dfrac{M}{m}}

\dfrac{D-x}{x}=\sqrt{\dfrac{m}{M}}

x=\dfrac{D}{1+\sqrt{\dfrac{m}{M}}}

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Slav-nsk [51]
<span>0.52% First, let's convert that speed into m/s. 150 km/h * 1000 m/km / 3600 s/h = 41.667 m/s Now let's see how much time gravity has to work on the ball. Divide the distance by the speed. 18 m / 41.667 m/s = 0.431996544 s Now multiply that time by the gravitational acceleration to see what the vertical component to the ball's speed that gravity adds. 0.431996544 s * 9.8 m/s^2 = 4.233566131 m/s Use the pythagorean theorem to get the new velocity of the ball. sqrt(41.667^2 + 4.234^2) = 41.882 m/s Finally, let's see what the difference is (41.882 - 41.667)/41.667 = 0.005159959 = 0.5159959% Rounding to 2 figures, gives 0.52%</span>
8 0
3 years ago
What ocean depth would the volume of an aluminium sphere be reduced by 0.10%
yKpoI14uk [10]

Answer:

6400 m

Explanation:

You need to use the bulk modulus, K:

K = ρ dP/dρ

where ρ is density and P is pressure

Since ρ is changing by very little, we can say:

K ≈ ρ ΔP/Δρ

Therefore, solving for ΔP:

ΔP = K Δρ / ρ

We can calculate K from Young's modulus (E) and Poisson's ratio (ν):

K = E / (3 (1 - 2ν))

Substituting:

ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)

Before compression:

ρ = m / V

After compression:

ρ+Δρ = m / (V - 0.001 V)

ρ+Δρ = m / (0.999 V)

ρ+Δρ = ρ / 0.999

1 + (Δρ/ρ) = 1 / 0.999

Δρ/ρ = (1 / 0.999) - 1

Δρ/ρ = 0.001 / 0.999

Given:

E = 69 GPa = 69×10⁹ Pa

ν = 0.32

ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)

ΔP = 64.0×10⁶ Pa

If we assume seawater density is constant at 1027 kg/m³, then:

ρgh = P

(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa

h = 6350 m

Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.

6 0
3 years ago
A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic
Elena L [17]

Answer:

the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

Explanation:

The torque is given by :

\bar {N} = \bar {m} * \bar {B}

where ;

m = 0.160 A.m²

B = 0.0800 T

θ = 35°

So the magnitude of the torque N = mBsinθ

N = (0.160)(0.0800)(sin 35°)

N = 0.007341

N = 7.34×10⁻³ Nm

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b) The potential energy \bar{U} = \bar{-m} * \bar{B}

U = -mBcosθ

U = (- 0.160)(0.0800)(cos 45)

U = -0.010485

U = -1.0485 ×10⁻² J

Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

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230×46=10580÷11=961 second
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