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Murrr4er [49]
2 years ago
7

Dos bolas, de masas mA = 40 g y mB =60 g, esta?n suspendidas como se observa en la figura. La bola ma?s ligera se jala en un a?n

gulo de 60° con respecto a la vertical y se libera. A) ?Cua?l es la velocidad de la bola ma?s ligera antes del impacto? B) ?Cua?l es la velocidad de cada bola despue?s de la colisio?n ela?stica? C) ?Cua?l sera? la altura ma?xima de cada bola despue?s de la colisio?n ela?stica?
Physics
1 answer:
Ymorist [56]2 years ago
3 0
MA=40 or something I really don’t know
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What type of System interact with its environment ​
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A 1000 kg car moving a 10 m/s collides with a stationary 2000 kg truck. The two vehicles interlock as a result of the collision.
IgorLugansk [536]

Answer:

v₃ = 3.33 [m/s]

Explanation:

This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.

In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

(m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}

where:

m₁ = mass of the car = 1000 [kg]

v₁ = velocity of the car = 10 [m/s]

m₂ = mass of the truck = 2000 [kg]

v₂ = velocity of the truck = 0 (stationary)

v₃ = velocity of the two vehicles after the collision [m/s].

Now replacing:

(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]

7 0
3 years ago
A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod
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(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
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(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
c_6 = 0.0812m+0.1625m=0.2437 m

(c) The total charge is Q=-3 \cdot 10^{-8}C. To get the charge on each piece, we should divide this value by 8, the number of pieces:
Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
q= -3.75\cdot 10^{-9}C
If we approximate piece 6 as a single  charge, the electric field is given by
E=k_e  \frac{q}{d^2}
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E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.
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Answer: Methane

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