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Jet001 [13]
4 years ago
9

n order better to map the surface features of the Moon, a 361 kg361 kg imaging satellite is put into circular orbit around the M

oon at an altitude of 147 km.147 km. Calculate the satellite's kinetic energy K,K, gravitational potential energy ????,U, and total orbital energy E.E. The radius and mass of the Moon are 1740 km1740 km and 7.36×1022 kg.
Physics
1 answer:
ollegr [7]4 years ago
3 0

Answer:

Explanation:

Mass of satellite

M_s = 361 kg

Distance of satellite from moon

h = 147 km = 147,000m

Radius of the moon is

R_m = 1740 km = 1740,000m

Mass of the moon is

M_m = 7.36 × 10²² kg.

The kinetic energy is equal to the potential energy of the body to the surface of the moon from the conservation of energy.

K.E = P.E = mgh

Gravity on moon is g = 1.62 m/s²

K.E = 361 × 1.62 × 147,000

K.E = 8.597 × 10^7 J.

B. The gravitational potential energy can be calculated using

U = G•M_s × M_m (1/R_s - 1 / R)

R is the total distance from the centre of the moon to the satellite

R = h + R_m = 147 + 1740 = 1887km

R = 1,887,000 m

U = 6.67 × 10^-11 × 361 × 7.36 × 10²² (1/1,740,000 - 1/1,887,000)

U = 6.67 × 10^-11 × 361 × 7.36 × 10²² × 4.48 × 10^-8

U = 7.93 × 10^7 J

Then,

The total energy becomes

E = K.E + U

E= 8.597 × 10^7 + 7.93 × 10^7 J

E = 1.653 × 10^8 J

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A charge Q acts on a point charge to create an electric field. Its strength, measured a distance of 40 cm away is 100 N/C. What
KIM [24]

Answer:

E_2=80N/C

Explanation:

From the question we are told that:

Initial Distance d_1=40cm=>0.4m

Initial Electric field strength E_1=100N/C

Final Distance d_2=20cm=>0.20m

Generally the equation for Electric field is mathematically given by.

 E=\frac{kq}{d^2}

 q=\frac{100*(0.4)^2}{K}

Substituting q for d=20cm

 E_2=\frac{k}{0.2}*\frac{100*(0.4)^2}{K}

 E_2=80N/C

6 0
3 years ago
True or false that humans emit light?
krok68 [10]

I'm pretty sure it's true. We glow with "Bioluminescence".

4 0
3 years ago
A que profundidad esta nadando una persona dentro de una alberca si la presión absoluta sobre ésta es de 156kPa?
lara31 [8.8K]

Answer:

La persona está nadando en la alberca a una profundida de 5.575 metros.

Explanation:

La presión absoluta (P_{tot}) experimentada por la persona es la suma de la presión atmosférica (P_{atm}) y la presión hidrostática de la columna de agua de la alberca (P_{h}), medidas en kilopascales. Es decir,

P_{tot} = P_{atm}+P_{h} (1)

P_{tot} = P_{atm} + \frac{\rho\cdot g \cdot z}{1000} (2)

Donde:

\rho - Densidad del fluido de la alberca, medida en kilogramos por metro cúbico.

g - Aceleración gravitacional, medida en metros por segundo al cuadrado.

z - Profundidad de la persona en la alberca, medida en metros.

Si sabemos que P_{atm} = 101.325\,kPa, \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} y P_{tot} = 156\,kPa, entonces la profundidad de la persona en la alberca es:

156 = 101.325 +\frac{(1000)\cdot (9.807)\cdot z}{1000}

54.675 = 9.807\cdot z

z = 5.575\,m

La persona está nadando en la alberca a una profundida de 5.575 metros.

5 0
3 years ago
HELP PLEASE
lisabon 2012 [21]

Sustainable buildings are generally less expensive over the lifetime of the building, due to lower energy costs in production of the materials and lower energy bills. Option A is correct.

<h3>What is sustainable buildings? </h3>

Sustainable buildings refer to both a structure and the implementation of environmentally responsible and resource-efficient methods.

Sustainable buildings are generally less expensive over the lifetime of the building, due to lower energy costs in production of the materials and lower energy bills,

Hence,option A is correct.

To learn more about the sustainable buildings, refer;

brainly.com/question/8434592

#SPJ1

5 0
2 years ago
Determine the average distance between the Earth and the Sun. Then calculate the average speed of the Earth in its orbit in kilo
FrozenT [24]

Answer:

The average speed of the Earth in its orbit is 4.74 km/s.

Explanation:

Distance between the Earth and the Sun ,r= 149.60 million km = 149.60\times 10^6 km

Considering the movement of Earth  around the Sun follows circular path.

The circumference of circular path will be give by :

D=2\pi r=2\times 3.14\times 149.60\times 10^6 km

Time taken by Earth to complete revolution around earth ,T= 1 year

1 year = 365 days

1 year = 31,536,000 seconds

T = 31,536,000 seconds

Speed of the Earth around the Sun :S

S=\frac{D}{T}=\frac{149.60\times 10^6 km}{31,536,000 s}

S=4.74 km/s

The average speed of the Earth in its orbit is 4.74 km/s.

8 0
3 years ago
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