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A certain planet has an escape speed . If another planet has twice size and twice the mass of the first planet, its escape speed

solong [7]

I think the correct answer from the choices listed above is option A. A certain planet has an escape speed . If another planet has twice size and twice the mass of the first planet, its escape speed will be Sqrt[2] V. Hope this answers the question.

3 0

3 years ago

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A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p

DedPeter [7]

Answer:

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

Explanation:

The knowable variables are

$d_{t}=25m$

$t_{1}=1.3 s$

$t_{2}=3.9 s$

$t_{3}=5.5 s$

Since the three traffic signs are equally spaced, the distance between each sign is $\frac{25}{3} m$

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

6 0

3 years ago

Mercury has an average disease to the sun of 0.39 AU. In two or more complete sentences, explain how to calculate the orbital pe

alukav5142 [94]

Answer: 88 Earth days

Explanation:

According to the Kepler Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit:

$T^{2}=a^{3}$ (1)

If we assume the orbit is circular and apply Newton's law of motion and the Universal Law of Gravity we have:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (2)

Where $M$ is the mass of the massive object and $G$ is the universal gravitation constant. If we assume $M$ constant and larger enough to consider $G$ really small, we can write a general form of this law:

$MT^{2}=a^{3}$ (3)

Where $T$ is in units of Earth years, $a$ is in AU (1 Astronomical Unit is the average distane between the Earth and the Sun) and $M$ is the mass of the central object in units of the mass of the Sun.

This means when we are making calculations with planets in our solar system $M=1$ .

Hnece, in the case of Mercury:

$(1)T^{2}=(0.39 AU)^{3}$ (4)

Isolating $T$ :

$T=\sqrt{(0.39 AU)^{3}}$ (5)

$T=0.243 Earth-years \frac{365 days}{1 Earth-year}=88.6 days \approx 88 days$ (6)

This means the period of Mercury is 88 days.

7 0

3 years ago

Which model could represent a neutral atom of nitrogen?

Semmy [17]

Answer:

In the Bohr model, a nitrogen atom has a central nucleus, composed of seven protons and seven neutrons, surrounded by seven electrons

Explanation:

thank me later

7 0

3 years ago

An automobile light has a 1.0-a current when it is connected to a 12-v battery. determine the resistance of the light.

Elis [28]

Resistance of automobile light is equal to 12 ohms.
Resistance = Voltage / Current
where
Voltage = 12 volts
Current = 1.0 Amperes
Resistance = 12 volts/ 1.0 amperes
Resistance = 12 ohms.
Automobile light has 12 ohms resistance when it is connected to 12volts battery with 1.0 ampere current.

3 0

3 years ago