Ask question

Ask question

All categories

3 years ago

10

In 1996, NASA performed an experiment called the Tethered Satellite experiment. In this experiment a 2.30 104-m length of wire w

as let out by the space shuttle Atlantis to generate a motional emf. The shuttle had an orbital speed of 7.50 103 m/s, and the magnitude of the earth's magnetic field at the location of the wire was 5.40 10-5 T. If the wire had moved perpendicular to the earth's magnetic field, what would have been the motional emf generated between the ends of the wire?

1 answer:

Nikolay [14]3 years ago

5 0

<h2>emf = 9.3 x 10³</h2>

Explanation:

When a conductor moves in the magnetic field , the emf is generated across its ends . Which can be calculated by the relation

emf ξ = B x l x v

here B is the magnetic field strength , l is the length of conductor and v is its velocity .

In our question B = 5.4 x 10⁻⁵ T

l = 2.30 x 10⁴ m and v = 7.5 x 10³

Thus ξ = 5.4 x 10⁻⁵ x 2.30 x 10⁴ x 7.5 x 10³ = 9.3 x 10³ Volt

You might be interested in

Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.40 kg and is initially moving to t

Dmitrij [34]

Answer:

The velocity of the first block is 1.15m/s while of the second block 2.56m/s.

Explanation:

Momentum is only conserved in an isolated system, and because this problem requires us to find the value of the two variables, we need two equations; therefore, to conserved momentum the energy must be released in to the system only after the collision has occurred.

Therefore, from conservation of momentum

$m_1v_1= m_1v_{1f}+m_2v_{2f}$

$0.4(2.4)= 0.4v_{1f}+0.5v_{2f}$

$\boxed{\:0.96= 0.4v_{1f}+0.5v_{2f}}$

and from conservation of energy

$\frac{1}{2}m_1v_1^2+1.2 = \frac{1}{2} m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2$

$m_1v_1^2+2.4 = m_1v_{1f}^2+m_2v_{2f}^2}$

$\boxed{\:3.804= 0.4v_{1f}^2+0.5v_{2f}^2.}$

Thus, we have two equations and two unknowns

$(1). \: 0.4v_{1f}+0.5v_{2f}=0.96$

${(2). \: 0.4v_{1f}^2+0.5v_{2f}^2 =3.804$

which has solutions

$(v_{1f}, v_{2f}) = (1.15,2.56)$

and

$(v_{1f}, v_{2f}) = (1.15,-2.56)$

Since the blocks cannot pass through each other, the 0.5kg block cannot have $v_{2f} = -2.56m/s$ (moves to the left) while the 0.4 kg block has $v_{1f} = 1.15m/s$ (moves to the right); therefore, we take the first solution for the velocities:

$(v_{1f}, v_{2f}) = (1.15,2.56)$ .

Thus , the velocity of the first block is 1.15m/s while of the second block 2.56m/s.

5 0

3 years ago

Read 2 more answers

A man drops a penny V=0 off the top of the Golden Gate Bridge how fast will the penny be moving when it hits the ground? The gol

BARSIC [14]

Answer:

Vf = 73.4 m/s

Explanation:

This is the case of vertical motion where we have to find the final velocity of the penny when it hits the ground. We can use 3rd equation of motion to find the final velocity:

2gh = Vf² - Vi²

where,

g = 9.8 m/s²

h = height = 275 m

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

2(9.8 m/s²)(275 m) = Vf² - (0 m/s)²

Vf = √5390 m²/s²

<u>Vf = 73.4 m/s</u>

7 0

3 years ago

Which types of light are absorbed by genetic material?

Aleksandr [31]

Explanation:

the answerer is visible

7 0

3 years ago

Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second

vampirchik [111]

To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

$I' = \frac{I_0}{2} cos^2 \theta$

Where

$\theta =$ Angle From vertical of the axis of the polarizing filter

$I_0 =$ Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

$I = \frac{I_0}{2}$

Replacing we have that

$I = \frac{370}{2}$

$I = 185W/m^2$

Re-arrange the equation,

$I'= \frac{I_0}{2}cos^2\theta$

Re-arrange to find \theta

$cos^2\theta = \frac{2I'}{I_0}$

$cos^2\theta = \frac{2*138}{370}$

$\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})$

$\theta = 0.5282rad$

$\theta = 30.27\°$

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

4 0

3 years ago

What should the Architect do to ensure Field-Level Security is enforced on a custom Visualforce page using the Standard Lead Con

OLEGan [10]

Answer:

the Architect should use {!$FieldType.lead.accessible} expression within the Visualforce page.

Explanation:

Visualforce is a framework that allows developers to build complex, user friendly interfaces that can be hosted primarily on the Lightning Platform

Controllers provide access to the data that should be displayed in a page, and can modify component behavior. a number of standard controllers are provided by The Lightning platform that contain functionality and logic that which are used for standard Salesforce pages

The Architect should Use the expression {!$FieldType.lead.accessible} within the Visualforce page.

3 0

3 years ago