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goldfiish [28.3K]

2 years ago

8

A typical mattress has a network of springs that provide support. If you sit on a mattress, the springs compress. A heavier pers

on compresses the springs more than a lighter person. Use the properties of springs and spring forces to explain why.

1 answer:

GenaCL600 [577]2 years ago

4 0

Answer:

Explanation:

Spring has a tendency to store energy in them and deform its shape when force is applied on it. Once the applied force is removed it regains its original shape and size.

It is in helical shape and is used in mattress to give structure and support. Spring have elastic nature and follows spring forces, F = k * x

where is the applied force, k is the spring constant and x is the amount of extension.

When a heavier person sits on a mattress, more weight is applied on springs and they form coils, as weight is removed they regains its shape again.

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(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500 - μC charge and flies due west at a sp

12345 [234]

(a) $2.64\cdot 10^{-8} N$ north

We can treat the aircraft as a single point charge moving in a magnetic field. In this case, the magnetic force exerted on the plane is

$F=qvB sin \theta$

where

$q=0.500 \mu C = 0.500\cdot 10^{-6} C$ is the charge on the plane

v = 660 m/s is the velocity

$B=8.00\cdot 10^{-5} T$ is the magnitude of the magnetic field

$\theta=90^{\circ}$ is the angle between the direction of motion of the jet and of the magnetic field

Substituting,

$F=(0.5\cdot 10^{-6})(660)(8.0\cdot 10^{-5})=2.64\cdot 10^{-8} N$

The direction can be found by using Fleming's left hand rule. We have:

- index finger: magnetic field direction (straight up)

- middle finger: velocity of the plane (due west)

- force: thumb --> north

(b) Not negligible

As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.

For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is

$W=mg=(80000)(9.8)=784,000 N$

As we can see, this is several orders of magnitude bigger than the magnetic force calculated at point (a), so the effects of the magnetic force are negligible.

8 0

2 years ago

A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to

White raven [17]

Answer:

<h2>The potential difference increases </h2>

Explanation:

from the relation $E= \frac{V}{d}$

where E= electric field (force per coulomb)

V= voltage

d= distance

Hence the voltage is going to be V= E×d.

Therefore this means that increasing the distance increases the voltage.

3 0

2 years ago

When light hits the boundary between two different materials, it can undergo when light hits the boundary between two different

Rashid [163]

When light hits the boundary between two different materials, it can undergo both reflection and refraction.

Reflection is the change in the direction of the wave that strikes the boundary between two materials.<span> It involves a change in the direction of waves when they clash with an obstacle.

Refraction involves the change in the direction of waves as they move from one medium to </span><span><span>another followed</span></span><span> by a change in speed and wavelength (this second medium should have different permitivity for the light to change its initial properties.)</span>

3 0

2 years ago

Read 2 more answers

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

Bezzdna [24]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?

0.57 m

0.64 m

0.80 m

1.25 m

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m (or 0.80 m)

_________________________________________

I Hope this helps, greetings ... Dexteright02! =)

8 0

2 years ago

Read 2 more answers

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a

Lapatulllka [165]

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that,

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :

$y = 0.049 \cos(7t)$ .......(1)

The general equation of a wave is given by :

$y=A\cos(\omega t)$ .......(2)

A is amplitude of wave

On comparing equation (1) and (2) we get :

A = 0.049 meters

So, the amplitude of the wave is 0.049 meters.

3 0

2 years ago