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3 years ago

Classify:Energyis the ability to exert force and cause change. Energy has manyforms:

Physics

1 answer:

Grace [21]3 years ago

6 0

Answer:

Energy: Kinetic, Sound, Thermal, gravitational, electric, light and nuclear

Explanation:

This exercise asks to clarify a little some important concepts

* Energy. It is the capable one that has the bodies to carry out a job.

* Kinetic energy. It is the energy due to the movement of objects

* sound energy. It is the energy due to the vibration of the atoms or molecules of the materials, it can be solid, liquid or gaseous

* Thermal energy. It is the energy due to movement of the particles

* Gravitational Energy. It is a configuration energy due to the height of the bodies with respect to a reference point

* Electric power. It is due to the movement of electric charges

* light energy. It is the energy that light carries due to the electric and magnetic fields that form it, it occurs throughout the electromagnetic spectrum

* Nuclear energy. It is the energy stored in the nucleus by the configuration and movement of protons and neutrons.

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The long handle on a rake is a _____.

artcher [175]

From what i know it is c. it is a lever

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3 years ago

A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The sprin

Mariana [72]

Answer:

The velocity of block = 0.188 $\frac{m}{s}$

Explanation:

Mass m = 5.6 kg

k = 1040 $\frac{N}{m}$

$\mu$ = 0.26

$x_{1} =$ 0.035 m , $v_{1}$ = 0

$x_{2} =$ 0.02 m

From work energy theorem

$K_{1} + U_{1} + W_{other} = K_{2} + U_{2}$ --------- (1)

Kinetic energy

$K = \frac{1}{2} k x^{2}$ ------- (1)

Potential energy

$U = \frac{1}{2} k x^{2}$ ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

$R_{N}$ = mg

$R_{N}$ = 5.6 × 9.81 = 54.9 N

Friction force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

$W_{f}$ = - 21.9 × 0.015

$W_{f}$ = - 0.329 J

Now kinetic energy

At point 1

$K_{1} = \frac{1}{2} m v^{2} _{1}$

$K_{1} = 0$

$U_{1} = \frac{1}{2} k x^{2}$

$U_{1} = \frac{1}{2} (1040) 0.035^{2}$

$U_{1} =$ 0.637 J

At point 2

$K_{2} = \frac{1}{2} (5.6) v^{2} _{2}$

$K_{2} = 2.8 v_{2} ^{2}$

Potential energy

$U_{2} = \frac{1}{2} k x_2^{2}$

$U_{2} = \frac{1}{2} (1040) 0.02^{2}$

$U_{2} = 0.208$ J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 $v_{2} ^{2}$ + 0.208

2.8 $v_{2} ^{2}$ = 0.1

$v_{2} =$ 0.188 $\frac{m}{s}$

This is the velocity of block.

6 0

4 years ago

Binding of a signaling molecule to which type of receptor leads directly to a change in the distribution of substances on opposi

Mrac [35]

When two sides of a membrane are in contact with each other, the distribution of ions will alter as a result of the binding of a signal molecule to a ligand-gated ion channel.

<h3>What is a ligand-gated ion channel?</h3>

Ligand-gated ion channels (LGICs) are membrane proteins that are structurally integral and feature a pore that permits the controlled passage of particular ions across the plasma membrane. The electrochemical gradient for the permeant ions drives the passive ion flux.

When a chemical ligand, such as a neurotransmitter, attaches to the protein, ligand-gated ion channels open. Changes in membrane potential cause voltage channels to open and close. When a receptor physically deforms, as in the case of pressure and touch receptors, mechanically-gated channels open.

Learn more about ligand-gated ion channel here:

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3 0

2 years ago

The cooking time for a roast scales like the 2/3rds power of the mass. Based on scaling laws, how much longer does a 20-lb roast

fgiga [73]

Answer:

2.726472 s more or 1.5874 times more time is taken than 10-lb roast.

Explanation:

Given:

- The cooking time t is related the mass of food m by:

t = m^(2/3)

- Mass of roast 1 m_1 = 20 lb

- Mass of roast 2 m_2 = 10 lb

Find:

how much longer does a 20-lb roast take than a 10-lb roast?

Solution:

- Compute the times for individual roasts using the given relation:

t_1 = (20)^(2/3) = 7.36806 s

t_2 = (10)^(2/3) = 4.641588 s

- Now take a ration of t_1 to t_2, to see how many times more time is taken by massive roast:

t_1 / t_2 = (20 / 10)^(2/3)

- Compute: t_1 / t_2 = 2^(2/3) = 1.5874 s

- Hence, a 20-lb roast takes 1.5874 times more seconds than 10- lb roast.

t_2 - t_1 = 2.726472 s more

3 0

3 years ago

If f3=0 and f1=12n, what does the magnitude of f⃗ 2 have to be for there to be rotational equilibrium?

QveST [7]

<span>I found out F2 and its correct just need help in solving the other parts </span>
<span>F2 = 4.5 N </span>

4 0

4 years ago