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zheka24 [161]
2 years ago
11

How is momentum conserved in a system in which two satellites connect?

Physics
2 answers:
Zielflug [23.3K]2 years ago
7 0

Answer:

The one satellite has all the momentum before they connect, and then afterwards they share it.

Explanation:

I took the test and I got it correct.

Hope this helps! :)

STALIN [3.7K]2 years ago
4 0
I think the correct answer from the choices listed above is the first option. The one satellite has all the momentum before they connect, and then afterwards they share it. <span>For a collision occurring between object 1 and object 2 in an isolated system, the total </span>momentum<span> of the two objects before the collision is equal to the total </span>momentum<span> of the two objects after the collision.</span>
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Un avión vuela a una velocidad de 900 km/h. Si tarda en viajar desde Canarias hasta la península 180 s ¿qué distancia recorre en
wel

Answer:

El avión recorrió 45 km en los 180 s.

Explanation:

La relación entre velocidad, distancia y tiempo se da de la siguiente manera;

Velocidad= \dfrac{Distancia}{Hora}

Por lo cual los parámetros dados son los siguientes;

Velocidad = 900 km/h = 250 m / s

Tiempo = 180 s

Estamos obligados a calcular la distancia recorrida

De la ecuación para la velocidad dada arriba, tenemos;

Distancia recorrida = Velocidad pf viaje × Tiempo de viaje

Distancia recorrida = 900 km/h × 180 s = 900

Distancia recorrida = 900 km/h × 1 h/60 min × 1 min/60 s × 180 s = 45 km

Por lo tanto, el avión viajó 45 km en 180 s.

8 0
3 years ago
How many volts would it take to push 1 amp through a resistance of 1 ohm?
ELEN [110]
V=I x R so V= 1 x 1 =1V
5 0
2 years ago
Read 2 more answers
A weight trainer lifts a 90.0-kg barbell from a stand 0.90 m high and raises it to a height of 1.75 m. What is the increase in t
nirvana33 [79]

Answer:

∆PE = 749.7 J

At 0.9 m high, PE = 793.8 J

At 1.75 m high, PE = 1543.5 J

7 0
2 years ago
Using Figure 2, what is the momentum of Train Car A before the collision?
Bess [88]

Answer:

Option A. 180000 Kgm/s.

Explanation:

From the question given above, the following data were obtained:

For Train Car A:

Mass of train car A = 45000 Kg

Velocity of train car A = 4 m/s

Momentum of train car A =?

For Train Car B:

Mass of train car B = 45000 Kg

Velocity of train car B = 0 m/s

Momentum is simply defined as the product of mass and velocity. Mathematically, it can be expressed as:

Momentum = mass × velocity

With the above formula, the momentum of train car A before collision can be obtained as follow:

Mass of train car A = 45000 Kg

Velocity of train car A = 4 m/s

Momentum of train car A =?

Momentum = mass × velocity

Momentum = 45000 × 4

Momentum of train car A = 180000 Kgm/s

5 0
2 years ago
A point charge q1=2.0μC is located on the positive y axis at y=0.30m, and an identical charge q2 is at the origin. Find the magn
dalvyx [7]

Answer:

(A) 0.279N at angle 38.02°

(B) 0.701N

(C) 14.19°

Explanation:

(A) The net force on q3 is given as:

F = Fxi + Fyj

Fx is the x component of the force

Fy is the y component of the force

Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0

Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx

First let us find y and angle x from the diagram.

Using Pythagoras theorem,

y² = 0.3² + 0.4²

y² = 0.25

y = 0.5m

Using SOHCAHTOA to find x,

sinx = 0.4/0.5

x = 53.13°

Electrostatic force, F is given as:

F = kqQ/r²

Where k = Coulumbs constant

F(1,3) = (k*q1*q3) / r²

F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)

F(1, 3) = 0.288N

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = 0.45N

Therefore,

Fx = -0.288cos36.87 + 0.45

Fx = 0.22N

Fy = 0.288cos53.13

Fy = 0.172N

=> F = 0.22i + 0.172j

The magnitude of the force will be

F(mag) = √(0.22² + 0.172²)

F(mag) = 0.279N

The direction of the force makes will be

tanθ = Fy/Fx

tanθ = 0.172/0.22 = 0.781

θ = 38.02° to the x axis.

(B) q2 = - 2.0 * 10^(-6)

This implies that:

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = -0.45N

Therefore,

Fx = -0.288cos36.87 - 0.45

Fx = -0.68N

Fy = 0.172N

=> F = - 0.68i + 0.172j

The magnitude of the force will be

F(mag) = √((-0.68)² + 0.172²)

F(mag) = 0.701N

(C) The direction of the force makes will be

tanθ = 0.172/0.68

θ = 14.19° to the x axis

4 0
3 years ago
Read 2 more answers
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