Stored energy and the energy of positions are _______________

A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a

ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω / t

(ω = √g/R)

∝ = $\frac{1}{t } \sqrt{\frac{g}{R} } }$

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = $\frac{2FR}{MR^2 + 2mr^2}$

Now we have ,

∝ = $\frac{1}{t} \sqrt{\frac{g}{R} }$ , ∝ = $\frac{2FR}{MR^2+2mr^2}$

equating both values

We have,

$F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }$

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

$F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N$

Each thruster has to applied a force of 294.5N in tangential direction

3 0

4 years ago

Compare and contrast electric potential energy and electric potential difference? Explain.

forsale [732]

Answer:

<u><em>Electric Potential Energy:</em></u>

The energy that is needed to move a charge against an electric firld is called Electric Potential Energy

<u><em>Electric Potential Difference:</em></u>

The amount of work done in carrying a unit charge from one point to an other in an electric field is called Electric Potential Difference.

<u><em>Relation:</em></u>

Relation between Electric potential and electrical potential energy is given by

$\delta V=\frac{PE}{q}$

Here PE represents Electric potential energy

and $\delta V$ is Electric potential difference

it means electric potential difference is the difference in electric potential energy divided by the charge.

6 0

3 years ago

(1) A positive charge +3 C is separated from another positive charge of +5 C by a distance of 7m. What is the magnitude of the e

Aneli [31]

1. The magnitude of the electric force between the two charges is 2.8×10⁹ N (Option B)

2. The net charge on the molecule is -8×10⁻¹⁹ C (Option D)

3. The magnitude of the force between the charges is 16000 N (Option C)

4. The correct statement is: A neutral object has equal numbers of protons and electrons. (Option C)

<h3>1. How to determine the force</h3>

Charge 1 (q₁) = +3 C
Charge 2 (q₂) = +5 C
Electric constant (K) = 9×10⁹ Nm²/C²
Distance apart (r) = 7 m
Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 3 × 5) / (7)²

F = 2.8×10⁹ N

<h3>2. How to determine the net charge on the molecule</h3>

Electron = 223 electrons
Proton = 218 protons
Net Charge =?

Charge = Proton - Electron

Charge = 218 - 223

Charge = -5 electrons

But

1 electron = 1.6×10⁻¹⁹ C

Thus,

Net Charge = -5 × 1.6×10⁻¹⁹ C

Net Charge = -8×10⁻¹⁹ C

<h3>3. How to determine the force</h3>

Charge 1 (q₁) = 2×10⁻⁴ C
Charge 2 (q₂) = 8×10⁻⁴ C
Electric constant (K) = 9×10⁹ Nm²/C²
Distance apart (r) = 0.3 m
Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 2×10⁻⁴ × 8×10⁻⁴) / (0.3)²

F = 16000 N

<h3>4. What is a neutral object?</h3>

A neutral object is an object having equal numbers of protons and electrons. For example, an object with 4 protons and 4 electrons is said to be neutral as illustrated below

Electron = 4 electrons
Proton = 4 protons
Net Charge =?

Charge = Proton - Electron

Charge = 4 - 4

Charge = 0 (neutral)

Thus, the correct statement about neutral object, given in the question is: A neutral object has equal numbers of protons and electrons (Option C)

Learn more about Coulomb's law:

brainly.com/question/506926

#SPJ1

6 0

2 years ago

A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th

Stolb23 [73]

Answer:

The current is $I_b = 400 \ A$