Which of the following pairs of charged particles will result in a repelling force acting between the particles?

The beat your doctor listens to through a sethoscope is the sound of the four values opening

djyliett [7]

and closing .

The heart has 4 valves. They are what makes the lub-dub lub-dub sounds that can be heard from the chest.

The mitral valve is located between the left atrium and the left ventricle. It closes the left atrium to collect oxygenated blood from the lungs and opens to pass it on to the left ventricle.

The tricuspid valve is located between the right atrium and the right ventricle. It closes the right atrium to hold unoxygenated blood and opens to pass it on to the right ventricle ensuring a one way flow.

The aortic valve is located between the aorta and the left ventricle. It closes the left ventricle and opens to the aorta to pass on the oxygen-rich blood to the body.

The pulmonary valve is located between the pulmonary artery and the right ventricle. It closes off the right ventricle and opens to pass on unoxygenated blood to the lungs.

5 0

3 years ago

From the average distance of one of jupiter's moons to jupiter and its orbital period around jupiter, we can determine:

den301095 [7]

Answer: Jupiter's mass

Explanation:

From Kepler's third law:

$T^2=\frac{4\pi^2}{GM}a^3$

where T is the orbital period of a satellite, a is the average distance of the satellite from the Planet, M is the mass of the planet, G is the gravitational constant.

If the average distance of one of Jupiter's moons to Jupiter and its orbital period around Jupiter is given then mass of the Jupiter can be found:

$\Rightarrow M_J=\frac{4\pi^2}{GT_m^2}a_m^3$

4 0

3 years ago

Read 2 more answers

Please help!!

IgorLugansk [536]

Answer:

Si un objeto se mueve en relación a un marco de referencia (por ejemplo, si una profesora se mueve a la derecha con respecto al pizarrón, o un pasajero se mueve hacia la parte trasera de un avión), entonces la posición del objeto cambia. A este cambio en la posición se le conoce como desplazamiento. La palabra desplazamiento implica que un objeto se movió, o se desplazó.

Explanation:

El desplazamiento se define como el cambio en la posición de un objeto. Se puede definir de manera matemática con la siguiente ecuación:

\text{desplazamiento}=\Delta x=x_f-x_0desplazamiento=Δx=x

f

−x

0

start text, d, e, s, p, l, a, z, a, m, i, e, n, t, o, end text, equals, delta, x, equals, x, start subscript, f, end subscript, minus, x, start subscript, 0, end subscript

x_fx

f

x, start subscript, f, end subscript se refiere al valor de la posición final.

x_0x

0

x, start subscript, 0, end subscript se refiere al valor de la posición inicial.

\Delta xΔxdelta, x es el símbolo que se usa para representar el desplazamiento.

Debemos ser cuidados al usar la palabra distancia, ya que hay dos maneras de usar el término en física. Podemos hablar acerca de la distancia entre dos puntos, o podemos hablar de la distancia recorrida por un objeto.

La distancia se define como la magnitud o el tamaño del desplazamiento entre dos posiciones. Observa que la distancia entre dos posiciones no es la misma que la distancia recorrida entre ellas.

Es importante darse cuenta que la distancia recorrida no tiene que ser igual a la magnitud del desplazamiento (es decir, la distancia entre dos puntos). De manera específica, si un objeto cambia de dirección en su trayecto, la distancia total recorrida será mayor que la magnitud del desplazamiento entre esos dos puntos. Ve los ejemplos resueltos a continuación.

8 0

3 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

3 years ago

Two cars are heading towards each other but are 12 km apart. one car is going 70 km/hr, and the other is going 50 km/hr. how muc

vova2212 [387]

12-50t=70t, t= 0.1h = 6 minutes.

3 0

3 years ago