Will mark brainlest helpp!!!!!!​

Answer:

may be d)45°

Explanation:

Answer:

two childrens are sitting on a see saw but can't swing.it means both weight is same (a=b)

Explanation:

Answer:

which subject is that question

Answer:

Weight = Mass × Gravitational field strength

Explanation:

The mass of the skydiver = 70 kg

The weight of the skydiver, W, is given as follows;

W = m × g

Where-

g = The gravitational field strength ≈ 9.81 m/s²

Therefore, the equation linking weight mass and gravitational field strength is presented as follows;

Weight = Mass × Gravitational field strength

Explanation:

hey

the answer is Brazil

hope it helps ✌

Answer:

enjoy

Explanation:

The number of time a cricket chirps in a minute is a function of the temperature in the formula: n=4t-160, where n=number of times a cricket chirps a minute and t=temperature in farenheit. You can use the temperature t=(1/4)n+40 to estimate the temperature based on the number of chirps per minute.

La respuesta correcta para esta pregunta abierta es la siguiente.

Te compartimos las tres ideas que te pueden ayudar para hacer tu artículo.

Título:

La Educación Física, una prioridad para las escuelas.

Existe una frase milenaria que se le atribuye a los griegos que dice "Mente sana en cuerpo sano."

Y es muy cierta.

Las escuelas deberían considerar seriamente aumentar el número de horas semanales para la impartición de la educación física por las siguientes razones.

1.- La educación física es vital para el desarrollo físico del estudiante. Desde los grados más básicos, las escuelas deberían fortalecer la enseñanza de la educación física por motivos de salud, y dejar este buen hábito en los alumnos para toda la vida.

2.- La educación física, mejora el rendimiento académico ya que ayuda oxigenando los músculos, fortaleciendo el cerebro para que pueda concentrase mejor, despeja a los alumnos de tal forma que puedan regresar a los salones e clase más "despiertos."

3.- La educación física sirve para introducir la importancia de practicar deporte toda la vida, y hacerlo en la etapa escolar a través de equipos deportivos como el Futbol Americano, el beisbol, el basquetbol, el volibol, el soccer, y otros tantos.

Este idea del trabajo en equipo a través del deporte es muy importante para desarrollar habilidades como el liderazgo, el compromiso, la constancia y el superar la adversidad.

El profesor de educación física debe enseñarte la forma correcta de respirar cuando haces ejercicio, la importancia de la relajación del cuerpo para iniciar una actividad. El profesor debe hacer énfasis en la etapa del calentamiento y el estiramiento para preparar al cuerpo antes de realizar el ejercicio. Así como muchas otras enseñanzas.

Por esa razón, las escuelas deberían considerar seriamente la posibilidad de aumentar las horas de educación física por semana.

Answer:

h = 2.78 m

Explanation:

Potential Energy(PE)

Mass(m)

Gravity(g) which is approximately equal to 10.

Height (h)

PE = mgh

41772.5 = 1500 * 10 * h

41772.5 = 15000h

h = 41772.5/15000

h = 2.78 m

It is a process of offering jobs to desired candidates and offering jobs to the desired applicant.

Answer:

The speed of the comet at the surface of the star is approximately 1,208,694.7 m/s

Explanation:

Question parameter obtained online; The mass of the star, M = 5 × 10³¹ kg

Explanation;

The given mass of the comet, m = 2 × 10⁸ kg

The initial velocity of the comet, v → 0

The distance of the comet from the star, d = 700,000,000 km

The gravitational potential at d = G·M·m/d

The kinetic energy of the comet, K.E. = m·v²/2

The kinetic energy of the comet at d = m·(0)²/2 = 0

The gravitational potential at the surface of the star, R = G·M·m/R

The kinetic energy of the comet at the surface of the star, R = m·(v)²/2 = 0

Where;

M = The mass of the star = 5 × 10³¹ kg

[tex]M_{Sun}[/tex] = The mass of the Sun = 1.989 × 10³⁰ kg

M/[tex]M_{Sun}[/tex] = 5 × 10³¹/(1.989 × 10³⁰) ≈ 25

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

R = The radius of the star

Therefore, we have;

m·(0)²/2 - G·M·m/d = m·v²/2 - G·M·m/R

∴ v = √((G·M·m/R - G·M·m/d)×2/m) = √(2·G·M(1/R - 1/d))

Therefore; v = (2 × 6.67430 × 10⁻¹¹ × 5 × 10³¹ × (1/R - 1/700,000,000,000))

v = 81696389149.1×√(1/R - 1/700,000,000,000).

The speed of the comet at the surface of the star, v = 81696389149.1×√(1/R - 1/700,000,000,000)

The mass radius relationship is given as follows;

[tex]\dfrac{R}{R_{Sun}} = 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]

[tex]R = R_{Sun} \times 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]

The radius of the Sun = 696,340,000 M

∴ R ≈ 696,340,000 × 1.3 × √(25.14) = 4538865694.76

R = 4538865694.76 m

v = 81696389149.1×√(1/4538865694.76 - 1/700,000,000,000) ≈ 1208694.7  m/s

Answer:

Ask with your Science teacher

Explanation:

i do not know the ans

Answer:

D

Explanation:

Cá xương, chim, thú, cá sấu không có sự pha trộn máu giàu O2 và máu giàu CO2 ở tim vì tim cá có 2 ngăn, tim các loài chim, thú, cá sấu có 4 ngăn

Answer:

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.

Explanation:

First, we need to determine the angular frequency of the system ([tex]\omega[/tex]), in radians per second:

[tex]\omega = \sqrt{\frac{k}{m} }[/tex] (1)

Where:

[tex]k[/tex] - Spring constant, in newtons per meter.

[tex]m[/tex] - Mass, in kilograms.

If we know that [tex]k = 10\,\frac{N}{m}[/tex] and [tex]m = 0.36\,kg[/tex], then the angular frequency of the system is:

[tex]\omega = \sqrt{\frac{10\,\frac{N}{m} }{0.36\,kg} }[/tex]

[tex]\omega \approx 5.270\,\frac{rad}{s}[/tex]

The kinematic formulas for the position ([tex]x(t)[/tex]), in meters, velocity ([tex]\dot x(t)[/tex]), in meters per second, and acceleration of the object ([tex]\ddot x(t)[/tex]), in meters per square second, are:

[tex]x(t) = A\cdot \cos \omega t[/tex] (2)

[tex]\dot x(t) = -\omega \cdot A \cdot \sin \omega t[/tex] (3)

[tex]\ddot x(t) = -\omega^{2}\cdot A \cdot \cos \omega t[/tex] (4)

Where [tex]A[/tex] is the amplitude of the motion, in meters.

From (2) we determine the time associated with position [tex]x(t) = 0.041\,m[/tex] ([tex]\omega \approx 5.270\,\frac{rad}{s}[/tex], [tex]A = 0.082\,m[/tex]):

[tex]t = \frac{1}{\omega}\cdot \cos^{-1} \left(\frac{x(t)}{A} \right)[/tex] (5)

[tex]t = \frac{1}{5.270\,\frac{rad}{s} }\cdot \cos^{-1}\left(\frac{0.041\,m}{0.082\,m} \right)[/tex]

[tex]t = 0.199\,s[/tex]

And the speed of the object is:

[tex]\dot x(t) = -\left(5.270\,\frac{rad}{s} \right)\cdot (0.082\,m)\cdot \sin \left[\left(5.270\,\frac{rad}{s} \right)\cdot (0.199\,s)\right][/tex]

[tex]\dot x(t) \approx -0.375\,\frac{m}{s}[/tex]

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.

Answer:

3.26 secs

Explanation:

Diameter of sphere ( D )= 10 mm

T1 = 75°C

P = 1 atm

T∞ = 23°C

T2 = 35°c

Velocity = 10 m/s

Determine how long it will take to cool the sphere to 35°C

Using the properties of copper and air given in the question

Nu = 2 + (Re)^0.8 (Pr)^0.33

hd / k = 2 + ( vd/v )^0.8 (Pr)^0.33

∴ h ≈  2594.7 W/m^2k

Given that :

(T2 - T∞) / ( T1 - T∞ )  = exp [ ( -hA / pv CP ) t ]  

( 35 - 23 ) / ( 75 - 23 ) = exp [  - 2594.7 * 6 * t / 8933 * 387 * 10 * 10^-3 ]

=  ln ( 12/52 ) = -1.466337069  =     - 0.45032919 *  t

∴ t ≈ 3.26 secs        ( -1.466337069 / -0.45032919 )

Answer:

Computer hard drives use magnetism to store the data on a rotating disk. More complex applications include: televisions, radios, microwave ovens, telephone systems, and computers. An industrial application of magnetic force is an electromagnetic crane that is used for lifting metal objects.

Answer:

Examples of magnetic force is a compass, a motor, the magnets that hold stuff on the refrigerator, train tracks, and new roller coasters. All moving charges give rise to a magnetic field and the charges that move through its regions, experience a force.

I Hope this will help you if not then sorry :)

Answer:

nước sôi: 440/17 l

nước 15°C:1260/17 l

Answer:

Amplitude.

Explanation:

A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.

In Science, there are two (2) types of wave and these include;

I. Electromagnetic waves: it doesn't require a medium for its propagation and as such can travel through an empty space or vacuum. An example of an electromagnetic wave is light.

II. Mechanical waves: it requires a medium for its propagation and as such can't travel through an empty space or vacuum. An example of a mechanical wave is sound.

An amplitude can be defined as a waveform that's measured from the center line (its origin or equilibrium position) to the bottom of a trough or top of a crest.

Hence, an amplitude is a word that describes the maximum displacement a point moves from its rest position when a wave passes.

On a graph, the vertical axis (y-axis) is the amplitude of a waveform and this simply means that, it's measured vertically.

Mathematically, the amplitude of a wave is given by the formula;

x = Asin(ωt + ϕ)

Where;

x is displacement of the wave measured in meters.

A is the amplitude.

ω is the angular frequency measured in rad/s.

t is the time period measured in seconds.

ϕ is the phase angle.

12 x cos 50 = 7.713451316...

Answer:

B. Two tuning forks that vibrate at the same frequency are near each other. One tuning fork is struck with a mallet so that it vibrates.

Explanation:

A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.

In Science, there are two (2) types of wave and these include;

I. Electromagnetic waves: it doesn't require a medium for its propagation and as such can travel through an empty space or vacuum. An example of an electromagnetic wave is light.

II. Mechanical waves: it requires a medium for its propagation and as such can't travel through an empty space or vacuum. An example of a mechanical wave is sound.

An amplitude can be defined as a waveform that's measured from the center line (its origin or equilibrium position) to the bottom of a trough or top of a crest.

Resonance can be defined as a phenomenon in which an object is forced to vibrate as a result of the vibration of a second object at the same natural frequency of the first object. Thus, it's an increase in the amplitude of oscillation that occurs when the frequency of a periodically applied force is equal to the natural frequency of the system on which it's acting.

Hence, the scenario that best describe a condition in which resonance can occur is two tuning forks that vibrate at the same frequency are near each other. One tuning fork is struck with a mallet so that it vibrates.

Answer:

B Two tuning forks that vibrate at the same frequency are near each other. One

tuning fork is struck with a mallet so that it vibrates.

Explanation: hope this helps ;)

Answer:

     i =[tex]- \frac{r \ A'}{2 \ rho}[/tex] ,  i =  0.92   A

Explanation:

This exercise asks for the electromotive force, which can be calculated with Faraday's law

          fem = [tex]- \frac{d \Phi_B }{dt}[/tex]

where the magnetic flux

          Ф = B. A

bold letters indicate vectors. We can write this equation

          Ф = B A cos θ

In this case the magnetic field is perpendicular to the page and the normal to the loop of the loop is also parallel to the page, therefore the angle is zero and the cosine is 1

the loop is

          A = π r²

we substitute in the first equation

          fem = - π r² [tex]\frac{dB}{dt}[/tex]

we substitute the values

          fem = -π r² 1

          fem = - π r²

to calculate the current let's use ohm's law

           V = i R

            R = ρ L / A'

where A 'is the area of ​​the wire and L is the length of the loop

            L = 2π r

            V = i (ρ 2π r / A ')

            I = [tex]\frac{V \ A'}{2\pi \ r \ rho}[/tex]

In this case

           V = fem

           I = fem / R            

           i =[tex]- \frac{r \ A'}{2 \ rho}[/tex]

In order to complete the calculation, you need the radius of the loop and / or the wire cutter.

if we assume that the loop has a radius of r = 1 cm = 0.01 m and an area of ​​the wire   A'= π 10⁻⁶ m²   a radius of the wire 1 mm

           i = - 10⁻² π 10⁻⁶ / ( 2 1.7 10-8)

           i =  0.92   A

Answer:

F1=26N and F2=09N ..this is from the two simultaneously equations

Answer:

Explanation:

By the time the hot water gets to the place where it can be used for heating, it will have lost a great deal of energy to the surroundings.

The way to prevent that from happening is to insulate the pipeline with fiber glass like house insulation. Of course since this is a physics question and not an engineering problem, you could move the power station closer to the houses to be heated.

The use of Insulating Layers with low Conduction Coefficients and radii greater than Critical Radius diminish Losses and gives important Savings in terms of Energy Efficiency.

The main problem of Transportation of Waste Energy through Long Distances is the heat transfer between the Fluid flowing through the Tube and Surroundings. Heat Losses are directly proportional to the Distance travelled by the Fluid.

A possible way to minimize this Problem is to isolate outer surface of Tubes with materials with low Conduction Coefficients. at Critical Radius, in which Heat Losses reach its theoretical Maximum, from which Heat Transfer tends to diminish at greater radii.

According to the Theory of Heat Transfer, the Critical Radius of the Insulating Layer is equal to:

[tex]r_{c} = \frac{k}{h}[/tex] (1)

Where:

Using materials with low Conduction Coefficient minimize Thickness requirements and, therefore, overall costs tends to sink when [tex]r > r_{c}[/tex].

Hence, the use of Insulating Layers with low Conduction Coefficients and Radii greater than Critical Radius diminish Losses and gives important Savings in terms of Energy Efficiency.

Answer:

I R 2

Explanat

Answer:

option C

Explanation:

Final velocity of the object is 114 m / s. Hence, final velocity of the object is 114 m / s.

Answer:

Answer:- Volume of the balloon is 5.78 L.

Solution:- There are 0.24 moles of hydrogen gas in a balloon at 35 degree C and 1.05 atm pressure. It asks to calculate the volume of the balloon.

This problem is based on ideal gas law equation:

P = 1.05 atm, n = 0.24 mole, T = 35 + 273 = 308 K

R =

V = ?

The equation could be rearranged for the volume as:

Let's plug in the values and do the calculations to get the volume of the balloon:

V = 5.78 L

So, the volume of the gas balloon is 5.78 L.

Answer:

the product of mass and velocity

....in my syllabus

Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

Answer:

The change in intensity is 63%.

Explanation:

intensity level = 18 db

Let the intensity is I.

Io = 10^(-12) W/m^2

Use the formula of intensity

[tex]dB = 10 log\left ( \frac{I}{Io} \right )\\\\18 = 10 log\left ( \frac{I}{Io} \right )\\\\1.8 = log\left ( \frac{I}{Io} \right )\\\\\left ( \frac{I}{Io} \right )=63.1[/tex]

So, the change in intensity is 63%.

Answer:

d. No, porque la ecuación de trabajo lo define.

Explanation:

En Física, el trabajo realizado se puede definir como la cantidad de energía transferida cuando un objeto o cuerpo se mueve a lo largo de una distancia debido a la acción de una fuerza externa.

Matemáticamente, el trabajo realizado viene dado por la fórmula;

[tex] W = F * d [/tex]

Dónde;

Por lo tanto, podemos deducir de la definición de trabajo y su fórmula que el trabajo se realiza cuando un objeto (cuerpo) se mueve una distancia o experimenta cualquier forma de desplazamiento mientras transfiere energía.

Answer:

I. Period = 1.3 seconds

II. Frequency = 0.769 Hertz

Explanation:

Given the following data;

Number of oscillation = 240 cycles

Time = 5.2 minutes.

Conversion:

1 minute = 60 seconds

5.2 minutes = X seconds

X = 60 * 5.2

X = 312 seconds

To find the following;

I. Period

Mathematically, the number of oscillation of a pendulum is given by the formula;

[tex] Number \; of \; oscillation = \frac {Time}{Period} [/tex]

Making period the subject of formula, we have;

[tex] Period = \frac {Time}{Number \; of \; oscillation} [/tex]

Substituting into the formula, we have;

[tex] Period = \frac {312}{240} [/tex]

Period = 1.3 seconds

II. Frequency

[tex] Frequency = \frac {1}{Period} [/tex]

Substituting the values into the formula, we have;

[tex] Frequency = \frac {1}{1.3} [/tex]

Frequency = 0.769 Hertz