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Luden [163]
3 years ago
15

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Earth's pull?
Physics
1 answer:
Tatiana [17]3 years ago
7 0

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = x+y=149.6\times 10^6\ m

G = Gravitational constant

M_s = Mass of Sun = 1.989\times 10^{30}

M_e = Mass of Earth = 5.972\times 10^{24}\ kg

According to the question

\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y

x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m

The probe should be 258774.9441 m from Earth

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What is the specific heat of a substance that absorbs 2500 joules of heat when a sample of 1.200 kg of the substance increases i
Anarel [89]

Answer:

0.035 J/g°C

Explanation:

From the question given above, the following data were obtained:

Heat (Q) absorbed = 2500 J

Mass (M) = 1.2 Kg

Initial Temperature (T₁) = 10 °C

Final Temperature (T₂) = 70 °C

Specific heat capacity (C) =?

Next, we shall determine the change in temperature. This can be obtained as follow:

Initial Temperature (T₁) = 10 °C

Final Temperature (T₂) = 70 °C

Change in temperature (ΔT) =

ΔT = T₂ – T₁

ΔT = 70 – 10

ΔT = 60 °C

Thus, the change in the temperature of the substance is 60 °C

Next, we shall convert 1.2 Kg to grams (g). This can be obtained as follow:

1 Kg = 1000 g

Therefore,

1.2 Kg = 1.2 Kg × 1000 g / 1 Kg

1.2 Kg = 1200 g

Thus, 1.2 Kg is equivalent to 1200 g.

Finally, we shall determine the specific heat capacity of substance. This can be obtained as follow:

Heat (Q) absorbed = 2500 J

Mass (M) = 1200 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) =?

Q = MCΔT

2500 = 1200 × C × 60

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3 years ago
Review. (c) Assume the dipole is a compass needle-a light bar magnet-with a magnetic moment of magnitude μ . It has moment of in
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The frequency of oscillation is \frac{1}{2\pi }  \sqrt{\frac{\mu B}{I} }.

<h3>What is a magnetic moment?</h3>

The magnetic moment is the magnetism of a magnet or other item that creates a magnetic field, as well as its orientation and strength. Electromagnets, permanent magnets, elementary particles like electrons, different compounds, and a variety of celestial objects are examples of things that have magnetic moments (such as many planets, some moons, stars, etc). The phrase "magnetic moment" typically refers to a system's magnetic dipole moment, which can be represented by an analogous magnetic dipole, which has a magnetic north and south pole that are barely separated from one another. For sufficiently small magnets or at sufficiently wide distances, the magnetic dipole component is adequate. For extended objects, additional terms may be required in addition to the dipole moment, such as the magnetic quadrupole moment.

T = \frac{ Id^{2}\theta }{dt^{2} }

-\mu B\theta=\frac{ Id^{2}\theta }{dt^{2} }

\frac{d^{2}\theta }{dt^{2} } = -(\frac{\mu BI}{\theta} )

By Comparing the above equation with the SHM equation

\frac{d^2 \theta} {dt^{2} } = -\omega^{2} \theta

\omega^{2} =\frac{ \mu B}{I}

Frequency = \frac{\mu}{2\pi }

=\frac{\sqrt{\frac{\mu B}{I} } }{2\pi}

=\frac{1}{2\pi }  \sqrt{\frac{\mu B}{I} }

To learn more about a magnetic moment, visit:

brainly.com/question/17000031

#SPJ4

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Answer:

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2 years ago
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