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3 years ago

What do you know about potential energy? And how do we use it?

1 answer:

7 0

Energy is the ability to do work or cause change. There are basically two main types of energy, kinetic and potential. Potential energy is energy that is stored. There are various types of stored, or potential energy. Chemical energy from a battery is a potential form of energy, elastic energy in a stretched rubber band is a form of potential energy, but the most commonly referred to form of potential energy in physics is that of gravitational potential energy. This is energy that is stored due to an object's position. It is dependent on the mass of the object, the height of the object above the ground or Earth, and the acceleration due to gravity.

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Explique ¿por que un objeto que tiene energía es capaz de realizar un trabajo?

Nana76 [90]

By striking another object that is free to move, the moving object can exert a force and cause the second object to shift its position. While the object is moving, it has the capacity for doing work. Energy means the ability to do work, so all moving things have energy by virtue of their motion.

Translation: Al golpear otro objeto que se puede mover libremente, el objeto en movimiento puede ejercer una fuerza y hacer que el segundo objeto cambie de posición. Mientras el objeto se mueve, tiene la capacidad de realizar un trabajo. Energía significa la capacidad de realizar un trabajo, por lo que todas las cosas en movimiento tienen energía en virtud de su movimiento.

3 0

2 years ago

A 6.00 cm tall light bulb is placed a distance of 54.2 cm from a double convex lens

BigorU [14]

Answer:

1 / f = 1 / i + 1 / o thin lens equation

1 / i = 1 / f - 1 / o = (o - f) / (o * f)

i = o * f / (o - f)

i = 54.2 * 12.7 / (54.2 - 12.7) = 16.6 cm image distance

Image is real and inverted and 16.6 / 54.2 * 6 = 1.94 cm tall

7 0

3 years ago

Where are you on Earth if you experience each of the following? (Refer to the discussion in Observing the Sky: The Birth of Astr

Aloiza [94]

Explanation:

We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.

Condition 1: The stars rise and set perpendicular to the horizon

The observer is at the equator

Condition 2: The stars circle the sky parallel to the horizon

The observer is at the Pole of the Earth

Condition 3: The celestial equator passes through the zenith

The observer is at the equator

Condition 4: In the course of a year, all stars are visible

The observer is at the equator

Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)

The observer is at North Pole

7 0

3 years ago

block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8

dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: $\sqrt{800} \,\,\,N$

Explanation:

First try to write all forces in vector component form:

The force F1 acting at 45 degrees would have multiplication factors of $\frac{\sqrt{2} }{2}$ on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is $F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

y-component of F1 is $F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

y-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is: $v_x=4\,*\,2=8\,m/s$