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ArbitrLikvidat [17]
3 years ago
12

Jack rolled a marble down a ramp and recorded the potential energy and kinetic energy of the marble at different positions on th

e ramp, as shown in the table. Marble experiment Height of Marble from Ground (meters) Potential Energy of Marble (Joules) Kinetic Energy of Marble (Joules) Total Energy of Marble (Joules) Trial 1 1 1 3 4 Trial 2 2 3 1 4 Which is the independent variable in this experiment? Mass of marble Kinetic energy of marble Potential energy of marble Height of marble from ground
Physics
1 answer:
mote1985 [20]3 years ago
6 0

Answer: D

Height of marble from ground

Explanation:

From the formula of kinetic energy and potential energy,

K.E = 1/2mv^2

While

P.E = mgh

From all the parameters given from the question. You can see that mass is constant, acceleration due to gravity is also constant.

Independent variable must be a value that can varies.

Since Jack rolled a marble down a ramp and recorded the potential energy and kinetic energy of the marble at different positions on the ramp to see the effects on both energies.

This different position must be the height which will produce an effect on the potential and kinetic energy of the marble.

Independent variable always provides an effect for dependent variable. Which are kinetic energy and potential energy in this case.

Height of marble from ground is the right answer.

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Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,
dsp73

Answer:

a) La magnitud del desplazamiento es de 5 m

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b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}

Lo que da;

\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}

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\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N

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La fuerza actúa a lo largo del plano.

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