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3 years ago

What are 20 examples of energy transformation?

1 answer:

8 0

1. Chemical energy from food is converted to mechanical energy when the food is broken down and absorbed in the muscles.
2.The chemical energy from food can also be converted to thermal energy to keep the body warm.
3.When lightning strikes a tree, electrical energy is converted to thermal energy.
4.Chemical energy stored within batteries can be converted to electrical energy.
5.Electrical energy can be converted to light energy when a light is switched out.
6.Electrical energy can also be converted to sound energy when it is used to power a loudspeaker.
7.Sound energy is converted to electric energy in a microphone.
8.At hydroelectric water plants, gravitational potential energy is converted to electrical energy when water falls from a height.
9.Wind turbines convert the mechanical energy from the wind into electrical energy.
10.Solar panels convert light energy into electrical energy.
11.Electrical energy can be converted to mechanical and sound energy in a blender.
12.Chemical energy is converted to mechanical energy in cars when gasoline undergoes combustion to power the engine.
13.When fuels such as gasoline are combusted, chemical energy is converted to heat and light energy.
14.Nuclear energy is converted to heat and light energy in a nuclear reaction as seen in the sun and in atomic bombs.
15.Plants convert light energy to chemical energy through photosynthesis. 16.During bungee jumps, gravitational potential energy is converted to elastic potential energy.
17.Friction converts kinetic energy to thermal energy.
18.Waves convert mechanical energy to electrical energy.
19.Geothermal energy in geologically active regions of the world can be harnessed as electrical energy.
20.A ball dropped from a height is an example of a change of energy from potential to kinetic energy.

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With a thunderstorm brewing, an electric field of magnitude 2.0 × 102 newtons/coulomb exists at a certain point in the earth’s a

Aleks04 [339]

Answer:

The correct option is (b).

Explanation:

Given that,

Electric field, $E=2\times 10^2\ N/C$

We need to find the magnitude of the force on the electron as a result of the electric field.

We know that, the electric force is given by :

$F=qE\\\\=1.6\times 10^{-19}\times 2\times 10^2\\\\F=3.2\times 10^{-17}\ N$

So, the required force on the electron is equal to $3.2\times 10^{-17}\ N$ .

5 0

3 years ago

at location a, what are the directions of the electric fields contributed by the electron. calculate the magnitudes of the elect

Lisa [10]

We can use the equation E = k | Q | r 2 E = k | Q | r2 to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge,

<h3>What is electric and magnetic field ?</h3>

With the use of electricity and other types of artificial and natural illumination, invisible energy fields known as electric and magnetic fields (EMFs) and radiation are created.

While the magnetic field is discernible by the force it exerts on other magnetic particles and moving electric charges, the electric field is actually the force per unit charge experienced by a non-moving point charge at any given location inside the field.

Learn more about Electromagnetic field here:

brainly.com/question/14372859

#SPJ4

4 0

1 year ago

Give two mathematical examples of Newton's third law and how you get the solution

bagirrra123 [75]

Answer:

1) Any particle moving in a horizontal plane slowed by friction, deceleration = 32 μ

2) The particle moving by acceleration = P/m - 32μ OR The external force = ma + 32μm

Explanation:

* Lets revise Newton’s Third Law:

- For every action there is a reaction, equal in magnitude and opposite

in direction.

- Examples:

# 1) A particle moving freely against friction in a horizontal plane

- When no external forces acts on the particle, then its equation of

motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ No external force

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

and R is the normal reaction of the weight of the particle on the

surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = 0 - F

∴ 0 - F = mass × acceleration

- Substitute F by μR

∴ - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

of gravity

∴ - μ(mg) = ma ⇒ a is the acceleration of motion

- By divide both sides by m

∴ - μ(g) = a

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ a = - 32 μ

* Any particle moving in a horizontal plane slowed by friction,

deceleration = 32 μ

# 2) A particle moving under the action of an external force P in a

horizontal plane.

- When an external force P acts on the particle, then its equation

of motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ The external force = P

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

and R is the normal reaction of the weight of the particle on the

surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = P - F

∴ P - F = mass × acceleration

- Substitute F by μR

∴ P - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

of gravity

∴ P - μ(mg) = ma ⇒ a is the acceleration of motion

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ P - 32μm = ma ⇒ (1)

- divide both side by m

∴ a = (P - 32μm)/m ⇒ divide the 2 terms in the bracket by m

∴ a = P/m - 32μ

* The particle moving by acceleration = P/m - 32μ

- If you want to fin the external force P use equation (1)

∵ P - 32μm = ma ⇒ add 32μm to both sides

∴ P = ma + 32μm

* The external force = ma + 32μm

7 0

3 years ago

A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spe

saw5 [17]

Answer:

Mass of ion will be $22\times 10^{-13}kg$

Explanation:

We have given ion is triply charged that is $q=3\times 1.6\times 10^{-19}=4.8\times 10^{-19}C$

Radius r = 36 cm = 0.36 m

Velocity of the electron $v=7\times 10^6m/sec$

Magnetic field B = 0.55 T

We know that radius of the path is given by $r=\frac{mv}{qB}$

$m=\frac{rqB}{v}=\frac{0.36\times 4.8\times 10^{-19}\times 7\times 10^6}{0.55}=22\times 10^{-13}kg$

6 0

3 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

3 years ago