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Kobotan [32]
2 years ago
15

Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio

nal force on you?

Physics
2 answers:
Pachacha [2.7K]2 years ago
5 0
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg. 

F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N

Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m

Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356  N

Ratio = 0.356  N/589.18 N
<em>Ratio = 6.04</em>
skelet666 [1.2K]2 years ago
3 0

The ratio of gravitational force of the sun to the earth's gravitational force is about 1 : 1580

\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>

radius of Earth = Re = 6.4 × 10⁶ m

mass of the Earth = Me = 6.0 × 10²⁴ kg

distance from Sun to Earth = Rs = 147 × 10⁹ m

mass of the Sun = Ms = 2.0 × 10³⁰ kg

<u>Asked:</u>

ratio of gravitational force of the sun and the earth = ?

<u>Solution:</u>

F_s : F_e = G \frac{ m M_s }{R_s^2} : G \frac{ m M_e }{R_e^2}

F_s : F_e = \frac{ M_s }{R_s^2} : \frac{ M_e }{R_e^2}

F_s : F_e = \frac{ 2 \times 10^{30} }{(147 \times 10^9)^2} : \frac{ 6.0 \times 10^{24} }{ (6.4 \times 10^6)^2}

F_s : F_e \approx 1 : 1580

\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

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A 5 kg object slows from +6 m/s to +2 m/s. What is the impulse?
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Impulse on the object: -20 kg m/s

Explanation:

The impulse exerted on an object is a vector quantity equal to the product between the force exerted on the object and the time interval during which the force is applied.

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v is the final velocity

For the object in this problem, we have

m = 5 kg

u = +6 m/s

v = +2 m/s

Therefore, its impulse is:

I=(5)(+2-(+6))=-20 kg m/s

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3 years ago
Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou
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Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

  • i = current flow = 3 A
  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
  • V = terminal voltage of the battery
  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

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I think you need Graph to figure it out

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An athlete runs at a constant velocity of 5.2 m/s. What is the velocity of the athlete relative to the ground?
dimulka [17.4K]

The relative velocity of the athlete relative to the ground is 5.2 m/s

The given parameters;

constant velocity of the athlete, V = 5.2 m/s

let the velocity of the ground = Vg = 0

The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.

The athlete is the moving object in this question while the ground is stationary.

The relative velocity of the athlete relative to the ground is calculated as follows;

V/V_g = V - V_g  = 5.2 - 0 = 5.2  \ m/s

Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s

Learn more here: brainly.com/question/24430414

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