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3 years ago

10

Una persona A tiene cierta cantidad de masa y una persona B tiene la mitad de masa de la persona A ¿como es el peso de B respect

o al de A?

1 answer:

musickatia [10]3 years ago

5 0

Answer:

El peso de la persona B es la mitad del peso de la persona A.

Explanation:

El peso de la persona B puede calcularse con la siguiente ecuación:

$P_{B} = m_{B}g$ (1)

En donde:

$m_{B}$ : es la masa de la persona B

g: es la gravedad

Dado que la persona B tiene la mitad de la masa de la persona A, tenemos:

$m_{B} = \frac{m_{A}}{2}$ (2)

En donde:

$m_{A}$ : es la masa de la persona A

Al introducir la ecuación (2) en (1) nos queda:

$P_{B} = \frac{m_{A}}{2}g$ (3)

Sabemos que el peso de la persona A está dado por:

$P_{A} = m_{A}g$ (4)

Entonces, al introducir la ecuación (4) en (3) tenemos:

$P_{B} = \frac{P_{A}}{2}$

Por lo tanto, el peso de la persona B es la mitad del peso de la persona A.

Espero que te sea de utilidad!

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nika2105 [10]

Answer:

the energy transformed into internal energy is E fr = 1.57*10¹⁰ J

Explanation:

From energy conservation , the internal energy gained by friction should be equal to the loss of total energy of the satellite

Since the total energy of the satellite can be decomposed into kinetic and potential energy

E fr = ΔE = E₂-E₁

E₁= K₁+V₁ = 1/2*m*v₁² + m*g*h₁

E₂ = K₂+V₂= 1/2*m*v₂² + m*g*h₂

first, we can choose our reference state so that h₂=0 and h₁=h=500 km

second , we can calculate the approximate the inicial velocity as the velocity required for a stable circular orbit

g = v₁²/(h+R) → v₁² = g*(h+R)

as the velocity diminishes, h diminishes, falling into the earth

assuming the radius of the Earth as R= 6371 km , then

v₁² = g*(h+R) = 9.8 m/s² * (500 km+ 6371 km) *1000 m/km = 6.73 * 10⁷ (m/s)²

replacing values

E₁ = 1/2*m*v₁² + m*g*h₁ = 1/2* 500kg *6.73 * 10⁷ (m/s)² + 500kg* 9.8m/s² * 500 km = 1.67*10¹⁰ J

E₂= 1/2*m*v₂² + m*g*h₂ = 1/2* 500kg *(2000 m/s)² + 0 = 1*10⁹ J

therefore

E fr = ΔE = E₂-E₁ = 1.67*10¹⁰ J - 1*10⁹ J = 1.57*10¹⁰ J

5 0

3 years ago

The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of

kolbaska11 [484]

Answer:

a) A = 4.0 m , b) w = 3.0 rad / s , c) f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

x = A cos (wt + fi)

In the exercise we are told that the expression is

x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

A = 4.0 m

b) the frequency or angular velocity

w = 3.0 rad / s

c) angular velocity and frequency are related

w = 2π f

f = w / 2π

f = 3 / 2π

f = 0.477 Hz

d) the period

frequency and period are related

T = 1 / f

T = 1 / 0.477

T = 20.94 s

e) the phase constant

Ф = 0.10 rad

f) velocity is defined by

v = dx / dt

v = - A w sin (wt + Ф)

speed is maximum when sine is + -1

v = A w

v = 4 3

v = 12 m / s

g) the angular velocity is

w² = k / m

k = m w²

k = 1.2 3²

k = 10.8 N / m

h) the total energy of the oscillator is

Em = ½ k A²

Em = ½ 10.8 4²

Em = 43.2 J

i) the potential energy is

Ke = ½ k x²

for t = 0 x = 4 cos (0 + 0.1)

x = 3.98 m

j) kinetic energy

K = ½ m v²

for t = 00.1 ²

v = A w sin 0.10

v = 4 3 sin 0.10

v = 1.98 m / s

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The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

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3 years ago

In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur

Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

$3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)$

First, we isolate C from equation (2):

$2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)$

using this value of C from equation (3) in equation (1):

$3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}$

<u>D = - 0.57</u>

Put this value in equation (3), we get:

$C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\$

<u>C = 2.43</u>

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3 years ago

Suppose you add two vectors A and B . What relative direction between them produces the resultant with the greatest magnitude? W

Akimi4 [234]

Answer:

Same direction to produce maximum magnitude and opposite direction to produce minimum magnitude

Explanation:

Let a be the angle between vectors A and B. Generally when we add A to B, we can split A into 2 sub vectors, 1 parallel to B and the other perpendicular to B.

Also let A and B be the magnitude of vector A and B, respectively.

We have the parallel component after addition be

Acos(a) + B

And the perpendicular component after addition be

Asin(a)

The magnitude of the resulting vector would be

$\sqrt{(Acos(a) + B)^2 + (Asin(a))^2}$

$= \sqrt{A^2cos^2a + B^2 + 2ABcos(a) + A^2sin^2a}$

$= \sqrt{A^2(cos^2a + sin^2a) + B^2 + 2ABcos(a)}$

$= \sqrt{A^2 + B^2 + 2ABcos(a)}$

As A and B are fixed, the equation above is maximum when cos(a) = 1, meaning a = 0 degree and vector A and B are in the same direction, and minimum with cos(a) = -1, meaning a = 180 degree and vector A and B are in opposite direction.

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4 years ago