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1 year ago

A force of 570 N keeps a certain ideal spring stretched a distance of 0.900 m.

1 answer:

3 0

Hooke's law states that the elongation of a material is proportional to the applied stress within the elastic limits of the material.

Using Hooke's law

F = kx

570 = k * 0.9

k = 712.5 N/m

(a) energy at x = 0.8 m

U = 0.5 k x^2 = 0.5* 712.8* 0.8^2

U = 228 J

(b) U = 0.5 * 712.8* 0.09^2

U = 2.887 J

Elastic potential energy is the potential energy released as a result of the deformation of an elastic body. B. Spring extension is stored. This is equal to the work done to stretch the spring, which depends on the spring constant k and the stretched distance. The force exerted by the spring is the restoring force that helps return the spring to its equilibrium length.

Learn more about Potential energy here:- brainly.com/question/14427111

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Why do boys like boys

Llana [10]

Answer:

maybe they're gay maybe they act gay so they can be around there girl crush cause a lot of girls like to hang out around gay guys

Explanation:

4 0

3 years ago

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Stels [109]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now by integrating above equation

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

4 0

3 years ago

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant accele

EleoNora [17]

Answer:

306.8264448 m

47.0016 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled by car

$s_c=ut+\dfrac{1}{2}at^2\\\Rightarrow s_c=\dfrac{1}{2}at^2$

Distance traveled by truck

$s_t=ut$

In order to overtake both distances should be equal

$\dfrac{1}{2}at^2=ut\\\Rightarrow \dfrac{1}{2}at=u\\\Rightarrow t=\dfrac{2u}{a}\\\Rightarrow t=\dfrac{2\times 23.5}{3.6}\\\Rightarrow t=13.056\ s$

$s_c=\dfrac{1}{2}at^2\\\Rightarrow s_c=\dfrac{1}{2}3.6\times 13.056^2\\\Rightarrow s_c=306.8264448\ m$

The distance the car has to travel is 306.8264448 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 3.6\times 306.8264448+0^2}\\\Rightarrow v=47.0016\ m/s$

The speed of the car when it overtakes the truck is 47.0016 m/s

8 0

3 years ago

Glider A of mass 2.5 kg moves with speed 1.7 m/s on a horizontal rail without friction. It collides elastically with glider B of

omeli [17]

Answer:

The speed of glider A after the collision is 0 m/s

Explanation:

Hi there!

The two gliders collide elastically. That means that the kinetic energy and momentum of the system are conserved, i.e., they remain constant before and after the collision.

The momentum is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass.

v = velocity.

The momentum of the system is the sum of the momentum of each glider.

be:

mA = mass of glider A

mB = mass of glider B

vA = velocity of glider A before the collision.

vB = velocity of glider B before the collision.

vA´= velocity of glider A after the collision.

vB´= velocity of glider B after the collision.

The momentum of the system will be:

mA · vA + mB · vB = mA · vA´ + mB · vB´

Replacing with the given data:

2.5 kg · 1.7 m/s + 2.5 kg · 0 m/s = 2.5 kg · vA´ + 2.5 kg · vB´

divide both sides of the equation by 2.5 kg:

1.7 m/s = vA´ + vB´

1.7 m/s - vA´ = vB´

Using the conservation of the kinetic energy of the system we can find vA´:

1/2 · mA · vA² + 1/2 · mB · vB² = 1/2 · mA · vA´² + 1/2 · mB · (1.7 m/s - vA´)²

Let´s replace with the given data:

1/2 · 2.5 kg · (1.7 m/s)² + 0 = 1/2 · 2.5 kg · vA´² + 1/2 · 2.5 kg · (1.7 m/s - vA´)²

divide both sides of the equation by (1/2 · 2.5 kg):

(1.7 m/s)² = vA´² + (1.7 m/s - vA´)²

(1.7 m/s)² = vA´² + (1.7 m/s)² - 2· 1.7 m/s · vA´ + vA´²

0 = 2vA´² - 2· 1.7 m/s · vA´

0 = 2vA´(vA´ - 1.7 m/s)

vA´ = 0

vA´ - 1.7 m/s = 0

vA´ = 1.7 m/s

Since the velocity of the glider A after the collision can´t be the same as before the collision, the velocity of glider A after the collision is 0 m/s.

4 0

3 years ago

If the area of the plates of a parallel plate capacitor is halved, and the separation between the plates tripled, all while the

Scorpion4ik [409]

Answer:

Explanation:

The capacitance of a parallel plate capacitor is given by

$C=\frac{\epsilon _{0}A}{d}$

where A is the area of plates and d is the separation between the plates.

Now the area is halved and the separation is tripled.

The new capacitance is

C' = C / 6

Initial potential energy is given by

$U=\frac{q^{2}}{2C}$

here the charge is constant

The new energy is given by

$U'=\frac{6q^{2}}{2C}$

U' = 6 U

The energy becomes six times the initial value.

3 0

3 years ago