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A spring has a force constant of 310.0 N/m. (a) Determine the potential energy stored in the spring when the spring is stretched

skad [1K]

Answer:

(a) 0.2618 J

(b) 0.1558 J

(c) 0 J

Explanation:

from Hook's Law,

The energy stored in a stretched spring = 1/2ke²

Ep = 1/2ke² ......................... Equation 1

Where k = spring constant, e = extension, E p = potential energy stored in the spring.

(a) When The spring is stretched to 4.11 cm,

Given: k = 310 N/m, e = 4.11 cm = 0.0411 m

Substituting these values into equation 1

Ep = 1/2(310)(0.0411)²

Ep = 155(0.0016892)

Ep =155×0.0016892

Ep = 0.2618 J.

(b) When the spring is stretched 3.17 cm

e = 3.17 cm = 0.0317 m.

Ep = 1/2(310)(0.0317)²

Ep = 155(0.0317)²

Ep = 155(0.0010049)

Ep = 0.155758 J

Ep ≈ 0.1558 J.

(c) When the spring is unstretched,

e = 0 m, k = 310 N/m

Ep = 1/2(310)(0)²

Ep = 0 J.

6 0

3 years ago

To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equil

Salsk061 [2.6K]

Answer:

The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N

Explanation:

As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.

Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.

So, the positions in our coordinate system are:

$r_1=(0,0)\\r_2=(4\ cos(60\°),4\ sin(60\°))\\r_3=(4,0)\\$

Now using Coulomb's force:

$F_{ij}=\frac{-kq_iq_j}{d^2}(\vv{r}_j-\vv{r}_i)$

Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:

Replacing we get 2 equations:

$F_{13}=\frac{-kq_1q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(-0.04\ cos(60\°),-0.04\ sin(60\°))\\\\F_{23}=\frac{-kq_2q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(0.04-0.04\ cos(60\°),-0.04\ sin(60\°))\\$

To work with the sam

F=∑F_i=3.5*10⁻⁴(0.023,0.032)+7*10⁻⁴(-0.016,0.032)=

=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=

F=(-5.6*10⁻⁶,3.36⁻⁵)N

7 0

3 years ago

Where is oceanic crust being formed?

Sever21 [200]

Oceanic crust is being formed it is because of the tectonic plates. Tectonic plates work because of volcanic eruption every time it will erupt the tectonic plates underground will have more pressure on magma, then below there will be a volcano thats why their is a volcano in ocean. The heat rises up then it will open, thats what we call divergent bounderies then their will be earthquakes form. Then another plate opens up and the crust is going down, it will stop if the volcano stops erupting because there is more lava left in volcano then it will go underwater and thats why oceanic crust is being formed.

#CarryOnLearning

Butbot na dili na tinuod ga timala ra ko ana

3 0

3 years ago

A mass of 4.8 kg is dropped from a height of 4.84 meters above a vertical spring anchored at its lower end to the floor. If the

krek1111 [17]

Answer:

45.6 cm

Explanation:

Let x (m) be the length that the spring is compressed. We know that when we drop the mass from 4.84 m above and compress the springi, ts gravitational energy shall be converted to spring potential energy due to the law of energy conservation

$E_g = E_p$

$mgh = 0.5kx^2$

where h = 4.84 + x is the distance from the dropping point the the compressed point, and k = 24N/cm = 2400N/m is the spring constant, g = 9.81 m/s2 is the gravitational acceleration constant. And m = 4.8 kg is the object mass.

$4.8*9.81(4.84 + x) = 0.5 * 2400 * x^2$

$47.088x + 227.906 = 1200x^2$

$1200x^2 - 47.088x - 227.906 = 0$

$x = 0.456m$ or 45.6 cm

5 0

4 years ago

A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis

Natali5045456 [20]

Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

$M_{before} = M_{after}$

where:

M = momentum [kg*m/s]

M = m*v

where:

m = mass [kg]

v = velocity [m/s]

$(m_{1} *v_{1} )-(m_{2} *v_{2})=(m_{1} *v_{3} )+(m_{2} *v_{4})$

where:

m1 = mass of the basketball = 0.5 [kg]

v1 = velocity of the basketball before the collision = 5 [m/s]

m2 = mass of the tennis ball = 0.05 [kg]

v2 = velocity of the tennis ball before the collision = - 30 [m/s]

v3 = velocity of the basketball after the collision [m/s]

v4 = velocity of the tennis ball after the collision = 34 [m/s]

Now replacing and solving:

(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)

1 - (0.05*34) = 0.5*v3

- 0.7 = 0.5*v

v = - 1.4 [m/s]

The negative sign means that the movement is towards left

3 0

3 years ago