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3 years ago

11

Occupants in a single space shuttle in orbit feel weightless. describe a scheme whereby occupants in a pair of shuttles (or even

one shuttle and another massive object in orbit) would be able to use a long cable to continuously experience a comfortably normal earthlike gravity.

1 answer:

agasfer [191]3 years ago

8 0

The scheme whereby occupants in a pair of shuttles is as follows
use a strong cable with large weight on the end
Then use the orbital naneuvering system(OMS) to set the whole work as spinning about their common center of gravity.

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Atoms in a molecule held together through sharing ______. (points 3)

Wittaler [7]

Covalent and ionic bonds

8 0

4 years ago

Read 2 more answers

Select the correct answer from each drop-down menu.

bagirrra123 [75]

Answer:

(1) An object that’s negatively charged has more electrons than protons.

(2) An object that’s positively charged has fewer electrons than protons.

(3) An object that’s not charged has the same number of electrons than protons.

Explanation :

Objects have three subatomic particles that are Electrons, protons, and neutrons.

Protons and neutrons are found in the nucleus and electrons rotate or move outside the nucleus. Naturally, protons are positively charged, neutrons have no charge, and electrons are negatively charged.

Therefore, an object that is negatively charged has more electrons than protons. An object that is not charged has the same number of electrons than protons. An object that is positively charged has fewer electrons than protons.

8 0

3 years ago

A situation known as ( blank)

Lisa [10]

Answer:

Apparent weightlessness

Explanation:

6 0

3 years ago

Read 2 more answers

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

b)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

c)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

4 years ago

A gas in a cylinder expands from a volume of 0.110 m³ to 0.320 m³. heat flows into the gas just rapidly enough to keep the press

Elden [556K]

80000 Joule is the change in the internal energy of the gas.

<h3>In Thermodynamics, work done by the gas during expansion at constant pressure:</h3>

ΔW = -pdV

ΔW = -pd (V₂ -V₁)

ΔW = - 1.65×10⁵ pa (0.320m³ - 0.110m³)

= - 0.35×10⁵ pa.m³

= - 35000 (N/m³)(m³)

= -35000 Nm

ΔW = -35000 Joule

Therefore, work done by the system = -35000 Joule

<h3>Change in the internal energy of the gas,</h3>

ΔV = ΔQ + ΔW

Given:

ΔQ = 1.15×10⁵ Joule

ΔW = -35000 Joule

ΔU = 1.15×10⁵ Joule - 35000 Joule

= 80000 Joule.

Therefore, the change in the internal energy of the gas= 80000 Joule.

Learn more about thermodynamics here:

brainly.com/question/14265296

#SPJ4

3 0

2 years ago