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Fynjy0 [20]
4 years ago
5

The gravitational force between Pluto and Charon is 3.61 × 1018 N. Pluto has a mass of 1.3 × 1022 kg, which is only slightly gre

ater than Charon’s mass of 1.6 × 1021 kg. How far apart are Pluto and Charon? 2.0 × 107 m 2.4 × 1012 m 3.8 × 1014 m 5.8 × 1024 m
Physics
2 answers:
raketka [301]4 years ago
4 0

<u>Answer:</u>

<em>The distance between Pluto and Charon is  1.96 \times 10^7  m</em>

<u>Explanation:</u>

Force of gravitation between two objects F_g= \frac{G(M_1 M_2 )}{r^2}  

Where M_1  \ and \ M_2  are the masses of the objects,r is the distance between the objects

G is the universal gravitational constant=6.67 \times 10^-^1^1 m^3/kg s^2

Here mass of pluto = 1.3 \times 10^2^2  kg

mass of charon = 1.6 \times 10^2^1 kg

Force of gravitation F_g=3.61 \times 10^1^8  N

F_g= \frac {G(M_1 M_2 )}{r^2}

r^2= \frac {G(M_1 M_2 )}{F_g }

r= \sqrt \frac{(G(M_1 M_2 )}{F_g}

=\sqrt \frac {((6.674\times 10^-^1^1) \times 1.3 \times 10^2^2 \times 1.6\times 10^2^1)}{(3.61 \times 10^1^8 )}

=\sqrt \frac {(13.88*10^32)}{(3.61*10^18)}

=1.96 \times10^7  m

chubhunter [2.5K]4 years ago
4 0

Answer:

The answer is A on edgen.

Explanation:

A. 2.0 × 10^7 m

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A small car of mass 833 kg is parked behind a small truck of mass 1767 kg on a level road. The brakes of both the car and the tr
Ymorist [56]

Answer: the acceleration of the truck a_t is 0.6911 m/s²

Explanation:

Given the data in the question;

There is no external force on the system; net force on the system is 0

Mass of the truck with the woman M = 1767 kg + 41 kg = 1808 kg

mass of the car m = 833 kg

car acceleration a_c = 1.5 m/s²

now let a_t be the acceleration of the truck in opposite direction

Action force on the car = Reaction force on the car

ma_c = Ma_t

a_t = ma_c / M

we substitute

a_t = (833 × 1.5) / 1808

a_t = 1249.5 / 1808

a_t = 0.6911 m/s²    

Therefore, the acceleration of the truck a_t is 0.6911 m/s²

8 0
3 years ago
Our Sun shines bright with a luminosity of 3.828x10^26 Watt. Her energy is responsible for many processes and the habitable temp
Sav [38]

Answer:

(a) 1317.44 W/m²

(b) 1.74×10¹⁵ litres of heating oil

(c) -20.63°C

(d) Energy storage in the Earth and the air

Explanation:

The parameters given are;

Luminosity of the Sun = 3.828 × 10²⁶ Watt

Distance of the Earth from the Sun, d = 152.06 × 10⁶ km

The radius of the Sun = 696 × 10³ km

The radius of the Earth, r_E = 6730 km

The surface area of the Sun = 12000 × Surface area of the Earth

The surface area of the Sun = 6.09 × 10¹² km²

Cross sectional area of the Earth = 1.27 × 10⁴ m² = 0.0127 km²

By the inverse square law, we have;

R = \dfrac{Luminosity \, of \, the \, Sun}{4 \pi d^2}

Where:

R = Solar radiation reaching the Earth

Therefore;

R = \dfrac{3.828 \times 10^{26}}{4 \times \pi \times (1.5206 \times 10^{11})^2} = 1317.44 \ W/m^2

Hence, the energy, E, reaching the Earth in a day is given as follows;

E = R × 4×π×r_E²×60×60×24 = 1317.44 × 4 × π × 6730000² × 60×60×24

E = 6.479×10²² Joules

(b) The number of litres of heating oil is therefore;

6.479×10²² J ÷ (37.3×10⁶ J/litre) = 1.74×10¹⁵ litres of heating oil

(c) 30% of the Energy is reflected, therefore;

0.7 × 6.479×10²² Joules = 4.54×10²² Joules reaches the Earths surface

From Stefan-Boltzmann law, we have;

T = \left (\dfrac{\left (1 - \alpha  \right ) \times R }{4\times \sigma }   \right )^{\dfrac{1}{4}}

Where:

α = 0.3

σ = Stefan-Boltzmann constant = 5.6704×10⁻⁸ W/(m²·K⁴)

Therefore;

T = \left (\dfrac{\left (1 - 0.3 \right ) \times 1317.44 }{4\times 5.6704 \times 10^{-8} }   \right )^{\dfrac{1}{4}} = 252.52 \ K = -20.63 ^{\circ} C

(d) The heat storage in the Earth and the air

7 0
3 years ago
Which combination of units expresses volume?<br> m?, cm², km?<br> m, cm, km<br> Om?cm. km
LenaWriter [7]

Answer:

Volume is often quantified numerically using the SI derived unit, the cubic metre.

Explanation:

7 0
3 years ago
Read 2 more answers
A 26 kg bin is stationary on the driveway. The coefficient of static friction is 0.25. You pull on the bin with a force of 52 N
konstantin123 [22]

The bin will not move because the frictional force is greater than the net force acting on the bin.

<h3>What is the net force acting on the bin?</h3>

The net force acting on the bin is obtained by subtracting the force acting on the Easterly direction from that acting in the Westerly direction.

Net force = 110 N - 52 N

Net force = 58 N

The frictional force acting on the bin is determined as well using the given formula:

Frictional force = coefficient of static friction * normal reaction

Norma reaction = 26 * 9.81

Normal reaction = 255.06 N

coefficient of static friction = 0.25

Frictional force = 0.25 * 255.06

Frictional force = 63.8 N

Since the frictional force is greater than the net force acting on the bin, the bin will not move.

In conclusion, the frictional force is the force that opposes relative between two objects at their surface of contact.

Learn more about frictional force at: brainly.com/question/13680415

#SPJ1

8 0
2 years ago
If an object is placed at a distance of 10 cm in front of a concave mirror of focal length 4 cm, find the position and character
Aleksandr [31]

Answer:

Explanation:

Focal length f = - 4 cm

Object distance u = - 10 cm

v , image distance = ?

Mirror formula

\frac{1}{v} +\frac{1}{u} = \frac{1}{f}

Putting the given values

\frac{1}{v} - \frac{1}{10} = - \frac{1}{4}

\frac{1}{v}=  - \frac{3}{20}

v = - 6.67 cm .

magnification

m = v / u

= - 6.67 / - 10

= .667

so image will be smaller in size in comparison with size of object .

Characteristics will be that ,

1 ) it will be inverted and

2 ) it will be real image .

4 0
3 years ago
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