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Marta_Voda [28]
2 years ago
3

1) The moon has a mass of 7.4 × 1022 kg and completes an orbit of radius 3.8 × 108 m about every 28 days. The Earth has a mass o

f 6 × 1024 kg and completes an orbit of radius 1.5 × 1011 m every year. What is the speed of the Moon in its orbit? Answer in units of m/s.
2) What is the kinetic energy of the Moon in orbit?
Answer in units of J.

3. What is the kinetic energy of the Earth? Answer in units of J.
Physics
2 answers:
Rus_ich [418]2 years ago
7 0
What is the speed of the Moon in its orbit? Answer in units of m / s.
 By definition we have that the speed in a circular movement is given by:
 v = w * r
 We then have to replace the values:
 v = ((2 * pi) / (28 * (24/1) * (3600/1))) * (3.8 * 10 ^ 8)
 v = 986.94 m / s

 2) What is the kinetic energy of the Moon in orbit? Answer in units of J.

 By definition, the kinetic energy is given by:
 K = (1/2) * (m) * (V ^ 2)
 Where,
 m: mass
 V: speed
 Substituting the values ​​we have:
 K = (1/2) * (7.4 * 10 ^ 22) * ((986.94) ^ 2)
 K = 3.60E + 28 J 

 3. What is the kinetic energy of the Earth? Answer in units of J. 
 First we must find the speed of the earth:
 v = ((2 * pi) / ((1) * (365/1) * (24/1) * (3600/1))) * (1.5 * 10 ^ 11)
 v = 29886 m / s
 By definition, the kinetic energy is given by:
 K = (1/2) * (m) * (V ^ 2)
 Where,
 m: mass
 V: speed
 Substituting the values ​​we have:
 K = (1/2) * (6 * 10 ^ 24) * ((29886) ^ 2)
 K = 2.67E + 33 J
professor190 [17]2 years ago
5 0
Mass of the moon Mm = 7.4 Ă— 10^22 kg 
Moon orbital radius Rm= 3.8 Ă— 10^8 m 
Time Tm = 28 days = 2.419 x 10^6
 Mass of the Earth Me= 6 Ă— 10^24 kg 
Earth orbital radius Re= 1.5 Ă— 10^11 m

 Time Te = 365 days = 3.154 x 10^7
 Length of Moon orbit Lm = 2 x pi x Rm = 2 x 3.14 x 3.8 Ă— 10^8 m = 23.86 x
10^8 m
 Moon orbital velocity Vm = Lm / Tm = 23.86 x 10^8 m / 2.419 x 10^6 = 986.5 m/s
 Kinetic Energy of the Moon = 1/2 x Mm x Vm^2 = 0.5 x 7.4 Ă— 10^22 x 986.5^2
 Kinetic Energy of the Moon = 36 x 10^27 J
 Length of earth orbit Le = 2 x pi x Re = 2 x 3.14 x 1.5 Ă— 10^11 = 9.42 x 10^11 m
 Earth orbital Velocity Ve = Lm / Tm = 9.42 x 10^11 / 3.154 x 10^7 = 29866 m/s
 Kinetic Energy of the Earth = 1/2 x Me x Ve^2 = 0.5 x 6 Ă— 10^24 x 29866^2
 Kinetic Energy of the Earth = 26.76 x 10^32 J

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REY [17]

Answer:

The magnitude of the velocity of the aircraft P relative to aircraft Q is zero

Explanation:

The velocity of the two aircraft, P & Q, v = 300 m/s

The angle of the direction between them, Ф = 90°

The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula

                                  <em> V = v cos Ф </em>

Substituting the values in the above equation

                                   v = 300 x cos 90°

                                      = 300 x 0

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Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.

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Answer: Anthony will be warmer after the game.

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The black color absorbs heat until a thermal equilibrium is attained. So, it is advisable to wear cotton clothes in summers not dark colored clothes.

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1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

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(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

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0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

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\boxed{v_0 = 28.58m/s.}

(b).

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v_x = 28.58m/s.

the vertical component of the velocity is

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2 years ago
A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled
SVETLANKA909090 [29]

Answer:

The  value is  v =  -0.04 \  m/s

Explanation:

From the question we are told that

   The  mass  of the block is  m  =  2.0 \ kg

   The  force constant  of the spring is  k  =  590 \ N/m

   The amplitude  is  A =  + 0.080

   The  time consider is  t =  0.10 \  s

Generally the angular velocity of this  block is mathematically represented as

      w =  \sqrt{\frac{k}{m} }

=>   w =  \sqrt{\frac{590}{2} }

=>   w = 17.18 \  rad/s

Given that the block undergoes simple harmonic motion the velocity is mathematically represented as  

         v  =  -A w sin (w* t )

=>       v  = -0.080 * 17.18 sin (17.18* 0.10 )

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