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Nostrana [21]
2 years ago
14

A glider of mass 0.240 kg is on a frictionless, horizontal track, attached to a horizontal spring of force constant 6.00 N/m. In

itially the spring (whose other end is fixed) is stretched by 0.100 m and the attached glider is moving at 0.400 m/s in the direction that causes the spring to stretch farther. What is the total mechanical energy (kinetic energy plus elastic potential energy) of the system?
Physics
1 answer:
Sveta_85 [38]2 years ago
6 0

Answer:

E_M=0.0492J.

Explanation:

The mechanical energy of the system will be the kinetic energy plus the elastic potential energy: E_M=K+U_e.

We know that the equation for the kinetic energy is K=\frac{mv^2}{2}, where <em>m </em>is the mass of the object and <em>v </em>its velocity.

We know that the equation for the elastic potential energy is U_e=\frac{k\Delta x^2}{2}, where <em>k</em> is the spring constant and \Delta x the compression (or elongation) respect to equilibrium.

So for our values we have:

E_M=K+U_e=\frac{mv^2}{2}+\frac{k \Delta x^2}{2}=\frac{(0.24kg)(0.4m/s)^2}{2}+\frac{(6N/m)(0.1m)^2}{2}=0.0492J.

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A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.
mote1985 [20]

Answer:

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let´s convert the spring constant units into N/m:

4 N/cm · 100 cm/m = 400 N/m

EPE = 1/2 · 400 N/m · (0.03 m)²

EPE = 0.18 J

When its length is 23 cm, the elastic potential energy of the spring is 0.18 J

b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:

EPE = 1/2 · 400 N/m · (0.06 m)²

EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

7 0
2 years ago
In the picture below, explain why the bear fell. Use FRICTION to explain your answer.
algol13
The bear fell because it slides to the surface of ice due to lack of friction.

One of these theories is that friction<span> causes the liquid layer of water to form on </span>ice<span>. </span>Friction<span> is the force that generates heat whenever two objects slide against each other. If you rub your hands together, you can feel them heat up. That's </span>friction<span> at work. When a </span>skate<span> moves over the surface of </span>ice, the friction<span> between the </span>skate<span> and the </span>ice<span> generates heat that melts the </span>outermost<span> layer of </span>ice<span>.</span>
5 0
2 years ago
The weight of an object is the same on two different planets. The mass of planet A is only sixty percent that of planet B. Find
natka813 [3]

Answer:

0.775

Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

F=G\frac{Mm}{R^2}

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

F_A=G\frac{M_Am}{R_A^2}

For planet B,

F_B=G\frac{M_Bm}{R_B^2}

We also know that the weight of the object on the two planets is the same, so

F_A = F_B

So we can write

G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}

We also know that the mass of planet A is only sixty percent that of planet B, so

M_A = 0.60 M_B

Substituting,

G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}

Now we can elimanate G, MB and m from the equation, and we get

\frac{0.60}{R_A^2}=\frac{1}{R_B^2}

So the ratio between the radii of the two planets is

\frac{R_A}{R_B}=\sqrt{0.60}=0.775

6 0
1 year ago
The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the e
Lapatulllka [165]

Answer:

0.2cm towards the retina.

Explanation:

the focal length of the frog eye is

(1/f) = (1/10) + (1/0.8)

f = 0.74cm

Since the distance of the object is 15cm Hence

(1/0.74) = (1/15) + (1/V)

V = 0.78cm

Therefore the distance the retina is to move is

0.78cm - 0.8cm = 0.02cm towards the retina.

3 0
2 years ago
You are learning about energy transformations in science class. Mel and Sam's built this set-up to see if light energy could be
Allisa [31]
I think it might be heat energy. light transforms into heat energy
5 0
1 year ago
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