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3 years ago

6

Here is a picture showing how two magnets will stick to one another when opposite poles align. When these magnets are pulled sli

ghtly apart, does the attractive force between them still exist?
A)Yes, but the new attractive force is due to gravity.
B)No, since they are not touching, no force is at work.
C)Yes, even though they are not touching, they are pulling on one another.
D)No, when the magnets are apart, they exert a repulsive force on one another.

2 answers:

7nadin3 [17]3 years ago

7 0

Answer:

C

Explanation:

A magnetic field exerts its force beyond just direct touch.

ser-zykov [4K]3 years ago

7 0

The correct answer is C!

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table 4.interaction between two permanent bar magnets what i did to the pair of magnets to cause interaction and observed the ef

Leni [432]

A magnet is a substance which attracts or repels another substance. In a magnet, the atoms are aligned in a particular direction in domains. A magnet has two poles: North pole and South pole. The domains are oppositely aligned in unlike poles. Like poles repel each other where as unlike poles attract each other. Hence, when we bring like poles closer, repulsion would be experienced. In case of unlike poles, they would stick together.

6 0

2 years ago

Read 2 more answers

Please someone help, I’m very confused and it’s due soon, thanks

Anit [1.1K]

Answer:

1 s
19.6 m
2 s
0.8 m/s^2
28 m/s
79 m/s
0.37 s
26 m/s
242 m/s
19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

PE = Mgh

At velocity v, the kinetic energy of the object is ...

KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

PE = KE

Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

2gh = v^2

This can be solved for height:

h = v^2/(2g) . . . . [eq1]

or for velocity:

v = √(2gh) . . . . [eq2]

__

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

t = 2(v^2/(2g))/v = v/g . . . . [eq3]

t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

g = 2h/t^2 . . . . [eq6]

__

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

__

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

__

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

__

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

__

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

__

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

__

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

2(9/49 s) ≈ 0.37 s

__

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

__

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

__

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

<em>Additional comment</em>

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

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3 years ago

What the dog doin? <br><br> Legend says, no one has ever known what the dog has been doing.

Xelga [282]

What dog lol, I need context pls

5 0

2 years ago

A transformer connected to a 120-v(rms) at line is to supply 13,000V(rms) for a neon sign.

lyudmila [28]

Answer:

(a) 108

(b) 110.500 kW

(c) 920.84 A

Solution:

As per the question:

Voltage at primary, $V_{p} = 120\ V$ (rms voltage)

Voltage at secondary, $V_{s} = 13000\ V$ (rms voltage)

Current in the secondary, $I_{s} = 8.50\ mA$

Now,

(a) The ratio of secondary to primary turns is given by the relation:

$\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}$

where

$N_{p}$ = No. of turns in primary

$N_{s}$ = No. of turns in secondary

$\frac{N_{s}}{N_{p}} = \frac{13000}{120}$ ≈ 108

(b) The power supplied to the line is given by:

Power, P = $V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW$

(c) The current rating that the fuse should have is given by:

$\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}$

$\frac{13000}{120} = \frac{I_{p}}{8.50}$

$I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A$

6 0

3 years ago

A solenoidal coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long

Airida [17]

Answer:

$0.00027646\ T$

$2.33\times 10^{-5}\ H$

-0.04194 V

Explanation:

$N_2$ = Number of turns in outer solenoid = 330

$N_1$ = Number of turns in inner solenoid = 22

$I_1$ = Current in inner solenoid = 0.14 A

$\dfrac{dI_2}{dt}$ = Rate of change of current = 1800 A/s

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

r = Radius = 0.0115 m

Magnetic field is given by

$B=\mu_0\dfrac{N_2}{l}I\\\Rightarrow B=4\pi \times 10^{-7}\times \dfrac{330}{0.21}\times 0.14\\\Rightarrow B=0.00027646\ T$

The average magnetic flux through each turn of the inner solenoid is $0.00027646\ T$

Magnetic flux is given by

$\phi=BA\\\Rightarrow \phi=0.00027646\times \pi 0.0115^2\\\Rightarrow \phi=1.14862\times 10^{-7}\ wb$

Mutual inductance is given by

$M=\dfrac{N_1\phi}{I}\\\Rightarrow M=\dfrac{22\times 1.14862\times 10^{-7}}{0.14}\\\Rightarrow M=2.33\times 10^{-5}\ H$

The mutual inductance of the two solenoids is $2.33\times 10^{-5}\ H$

Induced emf is given by

$\epsilon=-M\dfrac{dI_2}{dt}\\\Rightarrow \epsilon=-2.33\times 10^{-5}\times 1800\\\Rightarrow \epsilon=-0.04194\ V$

The emf induced in the outer solenoid by the changing current inthe inner solenoid is -0.04194 V

3 0

2 years ago