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Nostrana [21]
3 years ago
10

How much elastic potential energy is stored in a bungee cord with a spring constant of 10.0 N/m when the cord is stretched 2.00

m?
Physics
1 answer:
Aleonysh [2.5K]3 years ago
6 0

Answer:

<em> The elastic potential energy stored in the bungee cord = 20 J</em>

Explanation:

potential energy: This is the energy possessed by a body due to its position. The S.I unit of energy is Joules. The mathematical expression for elastic potential energy is given below

E = 1/2ke²................ Equation 1

Where E = elastic potential energy of the spring, k = force constant of the spring, e = extension

<em>Given: K = 10 N/m, e = 2.00 m</em>

<em>Substituting these values into Equation 1</em>

<em>E = 1/2(10)(2)²</em>

<em>E = 5×4</em>

<em>E = 20 Joules.</em>

<em>Therefore the elastic potential energy stored in the bungee cord = 20 J</em>

<em></em>

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"<em>F = m·A.</em>  The net force on an object is equal to the product of the object's mass and its acceleration."

The two statements say exactly the same thing. You can take either one and work out the other one from it, just by working with the units.

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Neurons that deliver sensory information from sensory receptors to the spinal cord are called __________.
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Answer:

First Order Neurons

Explanation:

First Order Neurons

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2 years ago
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the shock absorbers in a car act as a big spring with k= 21900 N/m. when a 92.5 kg person gets in, how far does the spring stret
r-ruslan [8.4K]

Answer: 0.04139m

Explanation:

First, we need to calculate the weight of the man which will be:

Weight = mass × acceleration due to gravity

Weight = mg

Weight = 92.5 × 9.8

Weight = 906.5N

Then, we calculate the force which will be:

F = kx

mg = kx

x = mg/k

x = 906.5/21900

x = 0.04139m.

The spring stretched for 0.04139m.

4 0
2 years ago
A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x
mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

Vf^2 = Vi^2 + 2*a*x

2*a*x = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

Let x = Xf -Xi

Where Xf is the final position of the cart and Xi the initial position of the cart.

x = 12.5 - 0

x = 12.5

The cart comes to a stop before changing direction

Vf = 0 m/s

a = (0^2 - 5^2)/ 2*12.5

a = - 1 m/s^2

The cart is decelerating

Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.

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3 years ago
A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
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