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SVEN [57.7K]
2 years ago
6

Four objects are situated along the y axis as follows: a 1.93-kg object is at +2.91 m, a 2.95-kg object is at +2.43 m, a 2.41-kg

object is at the origin, and a 3.99-kg object is at -0.496 m.
Where is the center of mass of these objects?
Physics
1 answer:
Anna [14]2 years ago
3 0

Answer:

0.958 m

Explanation:

So the total mass of the system is

M = 1.93 + 2.95 + 2.41 + 3.99 = 11.28  kg

let y be the distance from the center of mass to the origin. With the reference to the origin then we have the following equation

My = m_1y_1 + m_2y_2 +m_3y_3 + m_4y_4

11.28y = 1.93*2.91 + 2.95*2.43 + 2.41*0 + 3.99*(-0.496) = 10.806

y = \frac{10.806}{11.28} = 0.958 m

So the center of mass is 0.958 m from the origin

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two billiard balls moving along the same line hit each other head-on. each has a mass of 0.220 kg; one has an initial velocity o
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Hi there!

Since the collision is elastic, we must also satisfy the following condition:

Ei = Ef, or:

KEi = KEf

Begin by writing an expression for momentum. (p = mv) Remember that one ball's direction is negative; in this instance, we can let the second ball be moving LEFT.

mv1 + mv2 = mvf1 + mvf2

0.220(1.84) + 0.220(-.530) = 0.220(vf1 + vf2)

0.2882/0.220 = vf1 + vf2

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Now, we can express this as a conservation of energy:

1/2mv1² + 1/2mv2² = 1/2mvf1² + 1/2mvf2²

Plug in values and simplify:

0.403315 = 1/2m(vf1² + vf2²)

Simplify further:

3.6665 = vf1² + vf2²

Use the equation derived from momentum above and solve for one variable:

vf2 = 1.31 - vf1

Plug in this expression for vf2:

3.6665 = vf1² + (1.31 - vf1)²

Expand:

3.6665 = vf1² + 1.7161 - 2.62vf1 + vf1²

Simplify:

1.9504 = -2.62vf1 + 2vf1²

Solve for vf1 using a graphing calculator:

vf1 = -0.53 m/s or 1.84 m/s; we must figure out which one is correct.

Since v1 is heading to the right initially with a velocity of 1.84 m/s, we know that the ball's velocity could not have stayed the same in both magnitude and direction, so the final velocity must be -0.53 m/s.

Now, we can solve for the velocity of the other ball (initial of 0.53 m/s):

vf2 = 1.31 - (-0.53) = 1.84 m/s.

Now, you could have also made the connection that when two balls of the SAME MASS experience an ELASTIC collision, the velocities are simply "exchanged" from one to another. I just used this more "extensive" method to prove this.

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Consider two carts, of masses m and 2m, at rest on an air track. if you first push one cart for 3 s and then the other for the s
Marrrta [24]
F = m₁ a₁ = m₂ a₂

if m₁ = m and m₂ = 2m :

F = ma₁ = 2m  a₂ ⇒ a₁ = 2 a₂
 
since v = at + v₀  with t = 3, v₀ = 0 ⇒ v = 3a:

v₁ = 2 v₂

since p = vm with v₁ = 2v and v₂ = v :

p₁ = v₁m₁ = 2v ⁻ m
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8 0
3 years ago
A group of bike riders took a 4 hour trip. During the first 3 hours, they traveled a total of 50 km, but
zlopas [31]

Answer:

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The average speed for the entire trip is the sum of the total distance traveled divided by the total time of the trip.

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Now;

  Average speed  = \frac{total distance }{time taken}  

 Insert the parameters and solve;

 Average speed  = \frac{60km}{4hr}   = 15km/hr

7 0
3 years ago
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