Electrons are in constant motion around the nucleus because of

What causes air to become less dense and rise?

aalyn [17]

Answer:

That force is 'gravity'.

8 0

3 years ago

What formal regions are located in the western hemisphere

Tju [1.3M]

Answer: Part of Europe, part of Africa, part of Antarctic and the entire America.

Explanation: Western hemisphere or west hemisphere encompasses all regions located west of the longitude 0 °, or meridian of Greenwich.

The formal regions located in the western hemisphere is Europe, Africa, Antarctic and America.

4 0

3 years ago

A future use of space stations may be to provide hospitals for severely burned persons. It is very painful for a badly burned pe

inessss [21]

Answer:

1.5min

Explanation:

To solve the problem it is necessary to take into account the concepts related to Period and Centripetal Acceleration.

By definition centripetal acceleration is given by

$a_c = \frac{V^2}{r}$

Where,

V = Tangencial velocity

r = radius

With our values we know that

$a_c = \frac{V^2}{r}$

$\frac{V^2}{r} = \frac{1}{10}g$

Therefore solving to find V, we have:

$V = \sqrt{\frac{1}{10}g*r}$

$V = \sqrt{\frac{9.81*200}{10}}$

$V = 14m/s$

For definition we know that the Time to complete are revolution is given by

$t = \frac{Perimeter}{Speed}$

$t = \frac{2\pi R}{V}$

$t = \frac{2\pi * 200}{14}$

$t = 1.5min$

6 0

3 years ago

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

3 0

4 years ago

A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) calculate the average force

iragen [17]

(a) $-1.5\cdot 10^6 N$

First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:

$v^2 -u ^2 = 2ad$

where

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d = 1.00 cm = 0.01 m is the displacement of the person

Solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.01)}=-20000 m/s^2$

And the average force on the person is given by

$F=ma$

with m = 75.0 kg being the mass of the person. Substituting,

$F=(75)(-20000)=-1.5\cdot 10^6 N$

where the negative sign means the force is opposite to the direction of motion of the person.

b) $-1.0\cdot 10^5 N$

In this case,

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d = 15.00 cm = 0.15 m is the displacement of the person with the air bag

So the acceleration is

$a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.15)}=-1333 m/s^2$

So the average force on the person is

$F=ma=(75)(-1333)=-1.0\cdot 10^5 N$

7 0

4 years ago