Which is the main difference between cardio and strength training activities

CAN ANY OF YALL ANSWER DIS PLS!!!!!!!! I AM GIVING 20 POINTS BTW!!!!!!!

sveticcg [70]

PHYSICAL CHANGES :
Melting an ice cube.
Boiling water.
Mixing sand and water.
Breaking a glass.
CHEMICAL CHANGES :
Digesting food.
Cooking an egg.
Heating sugar to form caramel.

4 0

3 years ago

What scientist shot alpha particles into gold

laila [671]

The Geiger–Marsden experiment(s) (also called the Rutherford gold foil experiment) were a landmark series of experiments by which scientists discovered that every atom contains a nucleus where its positive charge and most of its mass are concentrated

5 0

3 years ago

Read 2 more answers

De manera individual subraya la respuesta correcta: A) El Método Científico de la física experimental y su búsqueda de respuesta

alexira [117]

Answer:

a. La ciencia.

Explicación:

El método científico de la física experimental y su búsqueda de respuestas tiene un principio básico que es la ciencia. La ciencia es la aplicación del conocimiento del mundo natural y social siguiendo una metodología sistemática basada en evidencias. Sin evidencias no hay aceptación de la observación ni se debe hacer teoría porque la evidencia es el principal criterio para aceptar o rechazar una hipótesis.

6 0

3 years ago

Light takes 8 minutes to reach the Earth, and the speed of light is 3.0×10^8 m/s. a) What is the orbital speed of the Earth arou

spin [16.1K]

Answer:

(a) 28690 m/s (b) $2.46x10^{33}J$

Explanation:

The orbital speed is define as:

$v = \frac{2 \pi r}{T}$ (1)

Where r is the radius of the trajectory and T is the orbital period.

To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.

The average velocity in a Uniform Rectilinear Motion is defined as:

$v = \frac{d}{t}$ (2)

Where v is the velocity, d is the covered distance and t is the time.

Equation 2 can be rewritten for d to get:

$d = vt$ (3)

In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.

The time will be converted to seconds so the units in equation 3 can match:

$8min . \frac{60s}{1min}$ ⇒ $480s$

$t = 480s$

Replacing all those values in equation 3 it is gotten:

$d = (3.0x10^{8}m/s)(480s)$

$d = 1.44x10^{11}m$

Kepler’s third law is defined as:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

$T = \sqrt{r^{3}}$

$T = \sqrt{(1.44x10^{11}m)^{3}}$

It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.

$T = \sqrt{1AU}$

$T = 1AU$

That can be expressed in units of years.

$T = 1AU . \frac{1year}{1AU}$

$T = 1year$

But there are 31536000 seconds in one year:

$T = 1year . \frac{31536000s}{1year}$

$T = 31536000s$

Finally, equation 1 can be used:

$v = \frac{2 \pi (1.44x10^{11}m)}{(31536000s)}$

$v = 28690 m/s$

So Earth orbital speed around the Sun is 28690 m/s.

b) What is its kinetic energy?

The kinetic energy is defined as:

$E = \frac{1}{2}mv^{2}$ (4)

Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:

$F = \frac{GMm}{r^{2}}$

$ma = \frac{GMm}{r^{2}}$ (5)

M will be isolated in equation 5:

$M = \frac{r^{2}a}{G}$

Where r is the radius of the Earth ( $6.38x10^{6}m$ )

$M = \frac{(6.38x10^{6}m)^{2}(9.8m/s^{2})}{6.67x10^{-11}kg.m/s^{2}.m^{2}/Kg^{2}})$

$M = 5.98x10^{24} Kg$

$E = \frac{1}{2}( 5.98x10^{24} Kg))(28.690m/s)^{2}$

$E = 2.46x10^{33}Kg.m^{2}/s^{2}$

$E = 2.46x10^{33}J$

Hence, the kinetic energy of Earth is $2.46x10^{33}J$ .

8 0

3 years ago

A typical human consumes 2500 Kcal of energy during a day. This is the equivalent to 10,450,000 J! Say you decided to run stairs

defon

Answer:

2940.1 joules would you burn in climbing stairs all day.

Explanation:

Work = W = F $\times$ d

going up stairs would be against force of gravity

W = mgh

where h is the height

the question is not complete because we need speed or distance

h = v $\times$ t

so assuming 1 step per second

h = 86,400 steps $\times$ 7inchs/step $\times$ 0.0254 m/inch

h = 15362 m

so from this

W = 800 N $\times$ 15362

= 12289600 J

that means YOU need 12289600 J to walk 1 step per second all day

divide that by 4180 J /Kcal

Kcal = $\frac{W}{(J/Kcal)}$

= $\frac{12289600}{4180}$

= 2940.1 Kcal

if you ran faster you would use more energy 2 steps per second would mean 5880 Kcal.

8 0

4 years ago