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Serjik [45]
2 years ago
12

2. For each pair of variables, identify which is the independent and which is the dependent variable. a. How much gas is left in

the gas tank vs. how far the car has traveled b. How much money you've spent vs. how much money is in your wallet c. How far a toy car traveled vs. how much time went by​
Physics
1 answer:
Korvikt [17]2 years ago
3 0

For each pair Independent variable and the dependent variable is -

a. How much gas is left in the gas tank vs. how far the car has traveled.

  • Independent variable =  how far the car has traveled
  • dependent variable = How much gas is left in the gas tank

b. How much money you've spent vs. how much money is in your wallet.

  • Independent variable =  How much money you've spent
  • dependent variable = how much money is in your wallet.

c. How far a toy car traveled vs. how much time went by​

  • Independent variable =  how much time went by​
  • dependent variable = How far a toy car traveled

An independent variable in any experiment or research is a variable that is manipulated or changed in the experiment, this change leads to a direct effect on the dependent variable.

A dependent variable is a variable that is directly affected by the independent variable and it is the variable that is measured or tested in an experiment.

Thus,

a. How much gas is left in the gas tank vs. how far the car has traveled.

  • Independent variable =  how far the car has traveled
  • dependent variable = How much gas is left in the gas tank

b. How much money you've spent vs. how much money is in your wallet.

  • Independent variable =  How much money you've spent
  • dependent variable = how much money is in your wallet.

c. How far a toy car traveled vs. how much time went by​

  • Independent variable =  how much time went by​
  • dependent variable = How far a toy car traveled

Learn more about dependent variables:

brainly.com/question/1670595:

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An object with total mass mtotal = 14.6 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.9 k
zheka24 [161]

Answer: 1) 0. 2) 4.2 Kg. 3) 15.4 m/s 4) 12.9 m/s 5) 0. 6) 3.62 KJ.

Explanation:

1) Assuming that no external forces act during the collision, total momentum must be conserved. As initially the total mass was at rest, so initial momentum is zero, final momentum of all the system must be 0 also.

2) After the explosion, as mass must be conserved also, the sum of the masses of the three pieces must be equal to the original total mass, so we can write the following:

m₁ + m₂ + m₃ = M = 14.6 Kg = 4.9 Kg + 5.5 Kg + m₃

Solving for m₃, we have:

m₃ = 14.6 Kg - 4.9 Kg -5.5 Kg = 4.2 Kg.

3) and 4)

As momentum is a vector, if it is magnitude must be 0, this means that all his components must be 0 too.

So, we can write two equations, one for the x-component, and other for the y-component, as follows:

pₓ = m₁. v₁ₓ + m₂.v₂ₓ + m₃.v₃ₓ = 0

py = m₁.v₁y + m₂. v₂y + m₃. v₃y =0

Replacing by the values, and solving for v₃ₓ and v₃y, we get:

v₃ₓ = 15.4 m/s

v₃y = 12.9 m/s

v = √(15.4)²+(12.9)² = 20.1 m/s

5) As the center of mass must move as if all the mass were concentrated in this point, and we know that the total momentum must be 0, this tells us that the magnitude of the velocity of the center of mass must be 0 too.

6) As initial kinetic energy is 0, as  the mass was at rest, the increase in the kinetic energy is obtained simply adding the kinetic energy of every piece of mass gained after explosion, as follows:

K = K₁ + K₂ + K₃ = 1/2 (m₁ . v₁² + m₂.v₂² + m₃.v₃²)

Replacing by the values, we get:

K= 3.62 KJ

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1.  C.  The change is easily reversible

2. A.  a physical change

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What is the weight of a 5.00 kg object on Earth? Assume g=9.81 m/s^2.
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<em>weight = (mass) x (gravity)</em>

Weight = (5.00 kg) x (9.81 m/s²)

weight = (5.00 x 9.81) (kg-m/s²)

<em>Weight = 49.05 Newton</em>

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