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Julli [10]
4 years ago
11

Two planets have the same surface gravity, but planet b has twice the radius of planet

Physics
2 answers:
Virty [35]4 years ago
8 0

The mass of planet B is 4m

\texttt{ }

<h3>Further explanation</h3>

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }

<em>F = Gravitational Force ( Newton )</em>

<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>

<em>m = Object's Mass ( kg )</em>

<em>R = Distance Between Objects ( m )</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

gravitational acceleration of Planet A = g₁ = g

gravitational acceleration of Planet B = g₂ = g

radius of Planet A = R₁ = R

radius of Planet B = R₂ = 2R

mass of Planet A = M₁ = m

<u>Asked:</u>

mass of Planet B = M₂ = ?

<u>Solution:</u>

We will compare the gravitational acceleration of the two planets as follows:

g_1 : g_2 = G\frac{M_1}{(R_1)^2} : G\frac{M_2}{(R_2)^2}

g_1 : g_2 = \frac{M_1}{(R_1)^2} : \frac{M_2}{(R_2)^2}

g : g = \frac{m}{(R)^2} : \frac{M_2}{(2R)^2}

1 : 1 = m : \frac{1}{4}M_2

\frac{1}{4}M_2 = m

\boxed{M_2 = 4m}

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

vampirchik [111]4 years ago
3 0
Planet A;
m = the mass 
Let r =  the radius 

Planet B:
Let M =  the mass
The radius is 2r (twice the radius of planet A)

The surface gravitational acceleration of planets A and B (they have the same surface gravity) are
g= \frac{Gm}{r^{2}} \, and \, g= \frac{GM}{(2r)^{2}} \\\\ m= \frac{M}{4} \\\\ M=4m

Answer: The mass of planet B is 4m.
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4 years ago
A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

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Height of sea water, z₁ = 10.5 m

gauge pressure, P_{gauge \ pressure} = 2.95 atm

Atmospheric pressure, P_{atm} = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

P₂ = P_{atm}

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

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RideAnS [48]

Answer:

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so we will have

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W = 187.2 J

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