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3 years ago

Which of the following scenarios will cause the current to stop flowing within a circuit?

2 answers:

8 0

C is correct. Only a closed circuit can carry current, opening a switch disconnects the circuit and stops current from flowing. Remember, all current flows in loops, if you can't make a loop out of it, current can't flow, and opening a switch, assuming no other paths for current exist, will completely stop the current.

Helga [31]3 years ago

7 0

The correct answer is C

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In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration ( $a_c$ ) and the tangential acceleration ( $a_t$ ):

$a=\sqrt{a_c^2+a_t^2}$

We know that:

a = 4.4g

$a_c = 3.2 g$

So, we can find the tangential acceleration:

$a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2$

The angular acceleration is related to the tangential acceleration by

$\alpha = \frac{a_t}{r}$

where r = 10.9 m is the length of the centrifuge. Substituting,

$\alpha = \frac{29.6}{10.9}=2.7 rad/s^2$

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8 0

3 years ago

State the difference between sensible heat and latent heat

Anna11 [10]

<u>Latent hea</u>t is related to changes in phase between liquids ,gases and solids.

<u>Sensible heat</u> is related to changes in temperature of a gas or object with no change in phase.

6 0

3 years ago

Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecul

pogonyaev

Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecule in a specific three-dimensional shape. The given statement is true.

<h3>What are hydrogen bonds?</h3>

A hydrogen bond is an electrostatic force of attraction among a hydrogen atom tightly attached to a more electronegative "donor" atom or group and another electronegative atom bearing a lone pair of electrons, known as the hydrogen bond acceptor.

Hydrogen bonds are too flimsy to connect atoms to form molecules, but they do hold various portions of a single large molecule together in a specific three-dimensional shape.

Thus, the given statement is true.

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5 0

1 year ago

In this experiment we will observe the magnetic fields produced by a current carrying wire. A long wire is suspended vertically,

Alisiya [41]

Answer:

See explanation

Explanation:

Solution:-

Electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass, as shown by a long straight current-carrying wire in. Hall probes can determine the magnitude of the field. Another version of the right hand rule emerges from this exploration and is valid for any current segment—point the thumb in the direction of the current, and the fingers curl in the direction of the magnetic field loops created by it.

Compasses placed near a long straight current-carrying wire indicate that field lines form circular loops centered on the wire. Right hand rule 2 states that, if the right hand thumb points in the direction of the current, the fingers curl in the direction of the field. This rule is consistent with the field mapped for the long straight wire and is valid for any current segment.

( See attachments )

- The equation for the magnetic field strength - B - (magnitude) produced by a long straight current-carrying wire is given by the Biot Savart Law:

$B = \frac{uo*I}{2\pi *r}$

Where,

I : The current,

r : The shortest distance to the wire,

uo : The permeability of free space. = 4π * 10^-7 T. m/A

- Since the wire is very long, the magnitude of the field depends only on distance from the wire r, not on position along the wire. This is one of the simplest cases to calculate the magnetic field strength - B - from a current.

- The magnetic field of a long straight wire has more implications than one might first suspect. Each segment of current produces a magnetic field like that of a long straight wire, and the total field of any shape current is the vector sum of the fields due to each segment. The formal statement of the direction and magnitude of the field due to each segment is called the Biot-Savart law. Integral calculus is needed to sum the field for an arbitrary shape current. The Biot-Savart law is written in its complete form as:

$B = \frac{uo*I}{4\pi }*\int\frac{dl xr}{r^2}$

Where the integral sums over,

1) The wire length where vector dl = direction of current (in or out of plane)

2) r is the distance between the location of dl and the location at which the magnetic field is being calculated

3) r^ is a unit vector in the direction of r.

3 0

3 years ago

Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to

jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

=K 2 +GmM (r 11− r 21)=6.9×10 7 J

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7 0

2 years ago