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katen-ka-za [31]
3 years ago
8

A spring is characterized by a spring constant of 60 N/m. How much potential energy does it store, when stretched by 1.0 cm?

Physics
2 answers:
Anestetic [448]3 years ago
8 0

Answer:

3.0 x10^-3 J

Explanation:

The potential energy of a spring is given by PE = (0.5)k*x^2  

Where

K: Spring Constant = 60 N/m

x: displacement of the spring from its equilibrium position = 1cm = 0.01m

Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J

fiasKO [112]3 years ago
5 0

Answer:

<em> The potential Energy stored in the spring = 0.003 J</em>

Explanation:

Potential Energy: This is the energy possessed by a body by virtue of its position. The ability to stretched or compressed an elastic material to do work is called the elastic potential energy. The unit of elastic potential energy is Joules (J).

Elastic potential energy can be represented mathematically as

W = 1/2ke² ............................... Equation 1

Where W = Elastic potential Energy, k = force constant of the spring. e = extension.

<em>Given: k = 60 N/m, e = 1.0 cm = (1/100) m = 0.01 m.</em>

<em>Substituting these values into equation 1,</em>

<em>W = 1/2(60)(0.01)²</em>

<em>W = 30×0.0001</em>

<em>W = 0.003 J.</em>

<em>T</em><em>hus potential Energy stored in the spring = 0.003 J</em>

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padilas [110]

The characteristics of the projectile launch allows to find the results for the questions about the movement of the ball are:

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  • In the movement is several dimensions, each one is independent of the others, the movement in the x axis does not affect the movement in the y axis.

Kinematics studies the motion of bodies looking for relationships between position, velocity and acceleration. In the case of vertical and projectiles launch  the acceleration on the vertical axis is the acceleration of gravity directed downward.

In the attachment we can see the position of the ball for two distances in the case of projectile launching.

We can see that the speed of the ball decreases with height according to the relation

          y = go t - ½ g t²

Where y is the height, g is the initial vertical velocity, g is the acceleration of gravity and t is time.

In all movements in various dimensions we assume that each movement in an . axis is independent

In the case of projectile launching, on the vertical axis there is an acceleration of gravity and on the horizontal axis there is no acceleration, the only parameter that this gives the two movements is the time, which is a scalar.

In conclusion, using the characteristics of the projectile launch, we can find the results for the questions about the movement of the ball are:  

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Learn more about projectile launch here: brainly.com/question/24888457

4 0
2 years ago
An aluminum-alloy rod has a length of 10.0 cm at 20°C and a length of 10.015 cm at the boiling point of water (1000C). (a) What
nikitadnepr [17]

Answer:

a.  9.99625 cm b. 68 °C

Explanation:

(a) What is the length of the rod at the freezing point of water (0 0C)?

Before we find the length of the rod, we need to find the coefficient of linear expansion, α = (L - L₀)/[L₀(T - T₀)] where L₀ = length of rod at temperature T₀ = 10.0 cm, T₀ = 20 °C, L = length of rod at temperature T = 10.015 cm and T = 100 °C

Substituting the values of the variables into the equation, we have

α = (L - L₀)/[L₀(T - T₀)]

α = (10.015 cm - 10.0 cm)/[10.0 cm(100 °C - 20 °C)]

α = 0.015 cm/[10.0 cm × 80 °C]

α = 0.015 cm/[800.0 cm °C]

α = 0.00001875 /°C

We now find the length L₁ at T₁ = 0 °C from

L₁ = L₀(1 + α(T₁ - T₀))

So, substituting the values of the variables into the equation, we have

L₁ = L₀(1 + α(T₁ - T₀))

L₁ = 10.0 cm[1 +  0.00001875 /°C(0° C - 20 °C)]

L₁ = 10.0 cm[1 +  0.00001875 /°C × -20° C]

L₁ = 10.0 cm[1 - 0.000375]

L₁ = 10.0 cm[0.999625]

L₁ = 9.99625 cm

(b) What is the temperature if the length of the rod is 10.009 cm?

With length L₃ = 10.009 cm at temperature T₃, using

L₃ = L₀(1 + α(T₃ - T₀))

making T₃ subject of the formula, we have

L₃/L₀ = 1 + α(T₃ - T₀)

L₃/L₀ - 1 = α(T₃ - T₀)

T₃ - T₀ = (L₃/L₀ - 1)/α

T₃ = T₀ + (L₃/L₀ - 1)/α

substituting the values of the variables into the equation, we have

T₃ = 20 °C + (10.009 cm/10.0 cm - 1)/0.00001875 /°C

T₃ = 20 °C + (1.0009 - 1)/0.00001875 /°C

T₃ = 20 °C + 0.0009/0.00001875 /°C

T₃ = 20 °C + 48 °C

T₃ = 68 °C

8 0
3 years ago
A computer base unit of mass 7.5 kg is dragged along a smooth desk. If the normal contact force is 23N and the tension in the ar
kramer
<span>If there isn't any force then the normal contact force will be 


N=m*g=7.5*9.81=73.58N 

which is 73.58-23=50.58N less 

so, there the person must pull at 23 degree upward 

break down the tension in two components, vertical and horizontal. 


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T=50.58/sin23=129.45N</span>
7 0
3 years ago
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777dan777 [17]
A geological fold<span> occurs when one or a stack of originally flat and planar surfaces, such as sedimentary strata, are bent or curved as a result of permanent deformation.

So A fold is a Bend? in a rock. Maybe.

</span>A fault<span> is a planar fracture or discontinuity in a volume of </span>rock<span>, across which there has been significant displacement as a result of </span>rock<span>-mass movement.</span>
3 0
3 years ago
Read 2 more answers
A certain parallel-plate capacitor is filled with a dielectric for which Κ = 5.5 .The area of each plate is 0.034 m2 , and the p
Nesterboy [21]

Answer:

The maximum energy that can be stored in the capacitor is  6.62 x 10⁻⁵ J

Explanation:

Given that,

dielectric constant k = 5.5

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separating distance, d =  2.0 mm = 2 x 10⁻³ m

magnitude of the electric field =  200 kN/C

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Maximum potential difference:

V = E x d

V = 200000 x 2 x 10⁻³ = 400 V

Maximum energy that can be stored in the capacitor:

E = ¹/₂CV²

E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²

E = 6.62 x 10⁻⁵ J

Therefore, the maximum energy that can be stored in the capacitor is  6.62 x 10⁻⁵ J

4 0
3 years ago
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