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3 years ago

Wha happens to the weight of astronauts when they are in orbit

Physics

2 answers:

Lynna [10]3 years ago

5 0

That depends on the size of the orbit.

In "low Earth orbit", like the Mercury, Gemini, early Apollo flights, and aboard the ISS, the astronauts' weight is about 85% of their weight when they're standing on the Earth's surface.

BUT ... since everything in orbit is in "free fall", they don't feel the weight of their own body or anything else. The feeling is like being completely without any weight.

Bingel [31]3 years ago

4 0

Microgravity. It is sometimes called zero- gravity, but this is wrong. What happens is they go into free fall. On earth, gravity causes all things to fall, it can essentially be called a 'vacuum' if you will. So, I'm not to sure about it, but that's kind of what happens.

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A cold glass of iced tea warms quickly on a hot day through the process of A. conversion. B. insulation. C. conduction. D. expan

r-ruslan [8.4K]

C) conduction. let me know if I'm wrong

7 0

4 years ago

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The Kyoto Protocol, which limits emissions of greenhouse gases for developed nations, is an example of which type of law? A. con

bazaltina [42]

The Kyoto Protocol, which limits emissions of greenhouse gases for developed nations, is an example of international Law

Explanation:

The Kyoto Protocol is named after a Japanese city in which it was executed in the year 1997 on 1st December.

The basic aim of the Kyoto Protocol is to reduce/Limit the emission of greenhouse gases which contributed to global warming

The Protocol was linked to the United Nations Framework Convention on Climate Change (UNFCCC)

The important factor that led to the failure of Kyoto's Protocol are deficiencies in the structure of the agreement, such as the exemption of developing countries from reductions requirements, or the lack of an effective emissions trading scheme

6 0

4 years ago

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A ball thrown horizontally at 12.6 m/s from the roof of a building lands 20.0 m from the base of the building

S_A_V [24]

Answer:

1.59 seconds

12.3 meters

but if you are wise you will read the entire answer.

Explanation:

This is a good question -- if not a bit unusual. You should try and understand the details. It will come in handy.

Time

Given

a = 0 This is the critical point. There is no horizontal acceleration.

d = 20 m

v = 12.6 m/s

Formula

d = vi * t + 1/2at^2

Solution

Since the acceleration is 0, the formula reduces to

d = vi * t

20 = 12.6 * t

t = 20 / 12.6

t = 1.59 seconds.

It takes 1.59 seconds to hit the ground

Height of the building

Givens

t = 1.59 sec

vi = 0 Another critical point. The beginning speed vertically is 0

a = 9.8 m/s^2 The acceleration is vertical.

Formula

d = vi*t + 1/2 a t^2

Solution

d = 1/2 a*t^2

d = 1/2 * 9.8 * 1.59^2

d = 12.3 meters.

The two vi's are not to be confused. The horizontal vi is a number other other 0 (in this case 12.6 m/s horizontally)

The other vi is a vertical speed. It is 0.

7 0

3 years ago

Is it possible to have a charge of 5 x 10-20 C? Why?

ruslelena [56]

1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is $2.3\cdot 10^{39}$ times bigger than the gravitational force

Explanation:

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

$e=1.6\cdot 10^{-19}C$

Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than $e$ , because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

$Q=5\cdot 10^{-20}C$

If we compare this value to $e$ , we notice that $Q$ , so no object can have a charge of $Q$ .

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

$Q=ne$

where

Q is the charge of the object

n is an integer multiple

e is the fundamental charge

Here we have

$Q=2.4\cdot 10^{-18}C$

Substituting the value of e, we find n:

$n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15$

n is integer, so this value of the charge is possible.

We now do the same procedure for the new object in this part, which has a charge of

$Q=2.0\cdot 10^{-19}C$

Again, the charge on this object can be written as

$Q=ne$

where

n is the number of electrons in the object

Using the value of the fundamental charge,

$e=1.6\cdot 10^{-19}C$

We find:

$n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25$

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

To solve this part, we use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (Action force), then object B exerts an equal and opposite force on object A (reaction force)".

In this problem, we have two objects:

- A charge Q

- A charge 5Q

Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

The magnitude of the electric force between two single-point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

r = 0.30 cm = 0.003 m is the separation

So, the electric force between the two charges is

$F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-6})(7.2\cdot 10^{-6})}{(0.003)^2}=32400 N$

The electric force between two charged objects has direction as follows:

- If the two objects have charges of opposite signs (+ and -), the force between them is attractive

- If the two objects have charges of same sign (++ or --), the force between them is repulsive

In this problem, the two charges are:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

We see that the two charges have same sign: therefore, the force between them is repulsive.

The electric force between the proton and the electron in the atom can be written as

$F_E=k\frac{q_1 q_2}{r^2}$

where

$q_1 = q_2 = e = 1.6\cdot 10^{-19}C$ is the magnitude of the charge of the proton and of the electron

$r=5.3\cdot 10^{-11} m$ is the separation between them

So the force can be rewritten as

$F_E=\frac{ke^2}{r^2}$

The gravitational force between the proton and the electron can be written as

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p = 1.67\cdot 10^{-27}kg$ is the proton mass

$m_e=9.11\cdot 10^{-27}kg$ is the electron mass

Comparing the 2 forces,

$\frac{F_E}{F_G}=\frac{ke^2}{Gm_p m_e}=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{(6.67\cdot 10^{-11})(1.67\cdot 10^{-27})(9.11\cdot 10^{-31})}=2.3\cdot 10^{39}$

8 0

3 years ago

A stream channel is formed when water

KonstantinChe [14]

A stream channel is formed when water enters in the water banks

7 0

3 years ago

Read 2 more answers