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Arte-miy333 [17]
3 years ago
13

Please help 50 points

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
7 0

Answer:

3a 0.7m/s

3b partially inelastic

4a 8.33 m/s

4b completely inelastic

5. puck: 10.59 m/s octopus: 10.59m/s

6. car: 117.44 m/s truck: 17.44m/s

7a - 1 m/s , the red cart travels to the left

7b) elastic

Explanation:

For all these questions, you have find momentum (P=mv) always remember initial P is always equal to final P

3.

Initial P:

mass of first ball x velocity of first ball + mass of second ball x velocity of second ball

0.5*3.5 + 0.5*0 = 1.75 kg.m/s

final P: also 1.75Kg.m/s

let x be the velocity for first ball

0.5*2.8+0.5*x=1.75

0.5x=1.75-1.4

     x=0.35/0.5

      x=0.7m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi does not equal to KEf

so the collision is partially inelastic, partially because the ball did not stick together

4.

Initial P:

2575*11 + 825*0 = 28,325 kg.m/s

final P: also 28,325 Kg.m/s

let x be the velocity for both vehicles

(2575+825)x=28325

                   x=28325/3400

                    x=8.33 m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi does not equal to KEf

so the collision is completely inelastic, completely because the vechicles slides off together

5

Initial P:

0.115*35 + 0.265*0 = 4.025 kg.m/s

final P: also 4.025Kg.m/s

since they slides off together, the velocity will be the same, so let x be the velocity for the puck and octopus,

(0.115+0.265)x=4.025

                     x=4.025/0.38

                     x=10.59

6. Initial P:

565*25+785*12 = 23,545 kg.m/s

final P: also 23,545 Kg.m/s

since they slides off together, the velocity will be the same, so let x be the velocity for the car and truck,

(565+785)x=23545

                 x=23545/1350

                 x=17.44

7.

Initial P:

0.25*2 + 0.75*0 = 0.5 kg.m/s

final P: also 0.5 Kg.m/s

let x be the velocity for red cart

0.75*1+0.25*x=0.5

0.25x=0.5-0.75

     x=-0.25/0.25

      x=-1m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi equals to KEf

so the collision is elastic

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Answer:

plz mark brainliest again lol :)

Explanation:

When you drop a ball from a greater height, it has more kinetic energy just before it hits the floor and stores more energy during the bounce—it dents farther as it comes to a stop.

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Answer:

the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

  top center: I2+I3 = 2.500 A

  bottom center: I2+I3-I1 = 0 A

  right: I3 = 1.875 A

Explanation:

You can write the KVL equations:

Top left loop:

  I2(4) +(I2 +I3)(2) +I1(1) = 10

Bottom left loop:

  (I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10

Right loop:

  (I2+I3)(2) +(I2+I3-I1)(2) = 5

In matrix form, the equations are ...

  \left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]

These equations have the solution ...

  \left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]

This means the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

  top center: I2+I3 = 2.500 A

  bottom center: I2+I3-I1 = 0 A

  right: I3 = 1.875 A

_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

  (I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

__

It helps to be familiar with the formulas for resistors in series and parallel.

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https://www.britannica.com/science/Newtons-laws-of-motion

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