All categories

3 years ago

How do you find the slope of a force-extension graph?

1 answer:

3 0

The relationship between force and extension is a linear one, which means that if you plot a force vs. extension graph, you'll get a straight line. It will pass through the origin (x = 0; F = 0), and its slope will be equal to the spring constant, k.

Measure the Slope of the Force Extension Graph

In general, you can find the slope of a line by choosing two points and forming a ratio of the rise and the run between these two points. If the first point you choose is (x1, F1), and the second point is (x2, F2), the slope of the line is:

slope= f(2)- f(1)

---------

x(2)-x(1)

Assuming F2 is larger than F1.

This is the value of the spring constant, k. Despite the minus sign in the Hooke's law equation, k is a positive number, because the slope in the Hooke's law graph is positive.

Note that the spring constant has units of force/distance. In the MKS system, the spring constant units are newtons/meter. In the CGS system, they are dynes/centimeter. In the imperial system, they are pounds of force (lbf) /foot.

Now that you have the spring constant, you can predict exactly how much the spring will distend or compress when you subject it to any force.

You might be interested in

Thermal energy is transferred between objects when they have?

Oliga [24]

The answer is B I had to answer this question one time

8 0

3 years ago

Read 2 more answers

How do sound waves travel?

Paul [167]

Sound waves travel in a wave pattern such as viberation by viberating it taking sound with it

6 0

3 years ago

A fish swims 10 cm from the front wall of an aquarium that is 35cm wide. The front wall of the aquarium is glass with negligible

DanielleElmas [232]

El problema es un caso generalmente tipico en optica concerniente a Apparent depth vs real depth

We see the objects closer than their real depth to the surface. We see objects only if the rays coming from them reaches our eyes.

The equation is given by,

$D_a = \frac{D_r}{\eta}$

Where,

$D_a =$ Apparenth depth

$D_r =$ Real depth

$\eta =$ Refractive index of the medium of object

For water $\eta$ is equal to 1.33

I attach an image of the theory that could help clarify the measurements.

We have,

$D_a = \frac{D_r}{\eta}$

$D_a = \frac{10}{1.333}$

$D_a = 7.5cm$

Therefore the apparent distance between the front wall of the aquarium to the fish is 7.5cm

B) The distance between fish and mirror is given by,

$d=35-10= 25$

So we have that real distance from the front wall of to image of fish is

$dr=25+35=60cm$

Applying our equation we have that,

$D_a = \frac{D_r}{\eta}$

$D_a = \frac{60}{1.333}$

$D_a = 45.1cm$

Therefore the apparent distance from the front wall of the aquarium to the image of the fish is 45.1cm

4 0

3 years ago

the term used to measure a change in speed of an object is a. velocity. b. acceleration. c. capacity. d. flow.

Tomtit [17]

The term is acceleration.

(b).

6 0

3 years ago

A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression a = (3.

mash [69]

Answer:

9.6 m

Explanation:

This is a case of motion under variable acceleration . So no law of motion formula will be applicable here. We shall have to integrate the given equation .

a = 3.6 t + 5.6

d²x / dt² = 3.6 t + 5.6

Integrating on both sides

dx /dt = 3.6 t² / 2 + 5.6 t + c

where c is a constant.

dx /dt = 1.8 t² + 5.6 t + c

when t = 0 , velocity dx /dt is zero

Putting these values in the equation above

0 = 0 +0 + c

c = 0

dx /dt = 1.8 t² + 5.6 t

Again integrating on both sides

x = 1.8 t³ / 3 + 5.6 x t² /2 + c₁

x = 0.6 t³ + 2.8 t² + c₁

when t =0, x = 0

c₁ = 0

x = 0.6 t³ + 2.8 t²

when t = 1.6

x = .6 x 1.6³ + 2.8 x 1.6²

= 2.4576 + 7.168

= 9.6256

9.6 m

5 0

3 years ago