In order for a wave to occur, there must be a _____________

How does the digestive system work with the blood vessels of the circulatory system?

igor_vitrenko [27]

Inhales, digestive system

7 0

3 years ago

Read 2 more answers

Enumerar 5 actividades en casa o en el trabajo en la que usualmente se mantenga una postura corporal incorrecta

ludmilkaskok [199]

Answer:

Existen cinco actividades en las que se mantiene una postura corporal incorrecta:

1) Sentarse en una silla.

2) Agacharse y levantar una objeto del piso, especialmente cuando es pesado.

3) Sentarse en un escritorio, especialmente frente a una computadora.

4) Llevar una mochila, especialmente si está sobrecargada.

5) Dormir sobre una cama en una posición inadecuada.

Explanation:

Existen cinco actividades en las que se mantiene una postura corporal incorrecta:

1) Sentarse en una silla.

2) Agacharse y levantar una objeto del piso, especialmente cuando es pesado.

3) Sentarse en un escritorio, especialmente frente a una computadora.

4) Llevar una mochila, especialmente si está sobrecargada.

5) Dormir sobre una cama en una posición inadecuada.

5 0

3 years ago

4 If someone gems to the Shore from about Kannada bhoot movies in the opposite direction .Explain, why?

dsp73

Hi Pupil Here's Your answer :::

➡➡➡➡➡➡➡➡➡➡➡➡➡

Answer : When a man jumps out from a boat, he pushes the boat with his feet (action) and the boat also exert an equal force on him in opposite direction reaction. As a result, the man jumps to the bank and the boat moves in the backward direction.

⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅

Hope this helps . . . . . .

8 0

3 years ago

A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

mafiozo [28]

Answer:

a. $f=1.22*10^{-15} N$

b. $f=53.6*10^{-17} N$

Explanation:

The force existing between two charges is given as

$f=\frac{kq_{1}q_{2}}{r^{2}}$

where q= charge,

k=constant

r= distance between the two charges

Note: this force can either be repulsive or attractive force depending on the charges involve. it is repulsive if they are similar charge and it is attractive if it is opposite charges.

Also the charge of an electron is

$-1.602*10^{-19}$

A. we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis

for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i$

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\ f=1.22*10^{-15} N$

b. at distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni$

this this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\ f=53.6*10^{-17} N$

8 0

3 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 15 m/sm/s. A passenger accidentally drops his camera

jeka94

To solve this problem we will apply the linear motion kinematic equations. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,

$h = v_0 t -\frac{1}{2} gt^2$