Which statement about magnetic poles is accurate?A) Same magnetic poles attract, while different magnetic poles repel.
B) Same magnetic poles attract, while different magnetic poles attract.
C) Opposite magnetic poles attract, while like magnetic poles repel.
D) Opposite magnetic poles repel, while like magnetic poles attract.

Answers

Answer 1

Answer:

The statement about magnetic poles, " Opposite magnetic poles attract, while like magnetic poles repel" is accurate. Hence, Option A is the correct answer.

What are Magnetic poles?

A Bar magnet has two poles i.e., the north pole and the south pole. The bar magnet has only two sides, the opposite poles doesn't repel each other but attracts each other while same pole repels one another.

The interaction between the magnets, the magnetic materials and all the charges is that the south magnetic pole attracts the north pole of another magnet.

When we consider the given options, As the same magnetic poles repels, Option A , B and D are wrong.

The statement about Magnetic poles, "Opposite magnetic poles attract, while like magnetic poles repel" is the correct statement.

Hence, Option C is the correct answer.

Learn more about Magnetic Poles,

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Answer 2

Answer:

C) Opposite magnetic poles attract, while like magnetic poles repel.

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At what distance from a long straight wire carrying acurrentof 5.0A is the magnitude of the magnetic field due to thewireequal to the strength of the Earth's magnetic field of about5.0 x10^-5 T?

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. the bicycle stops in 5.0 s. how far did the bicycle travel during the 5.0 seconds of braking?

Answers

Distance traveled by the bicycle during the 5 seconds of braking is 22m

Explanation:

initial angular velocity= 2 rev/s

final angular velocity= 0 rev/s

Angular displacement Ф= $((wi+wf)/(2) )$ t

Ф= $((0+2)/(2) )5=5$ rev

so the distance travelled= 5(2πr)

distance=5(2π*0.7)

distance=22m

The bicycle traveled about 22 m during the 5.0 seconds of braking

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Further explanation

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

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Centripetal Force can be formulated as follows:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

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Given:

radius of wheel = R = 0.70 m

initial angular speed = ω = 2.0 rev/s = 4π rad/s

final angular speed = ωo = 0 rad/s

time taken = t = 5.0 s

Asked:

distance covered = d = ?

Solution:

$d = \theta R$

$d = (\omega + \omega_o)(1)/(2)t R$

$d = ( 4 \pi + 0 ) (1)/(2)(5.0)( 0.70 )$

$d = 4\pi (1.75)$

$d = 7\pi \texttt{ m}$

$d \approx 22 \texttt{ m}$

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Learn more

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

Suppose a boat moves at 16.4 m/s relative to the water. If the boat is in a river with the current directed east at 2.70 m/s, what is the boat's speed relative to the ground when it is heading east, with the current, and west, against the current? (Enter your answers in m/s.)

Answers

Answer:

with the current: 19.1 m/s eastern

against the current: 13.7 m/s western

Explanation:

The boat speed relative to ground is the sum of the boat speed relative to water + the water speed relative to ground.

Suppose eastern is positive, water speed due east is 2.7m/s.

When the boat is heading east, its speed relative to water is 16.4m/s, its speed relative to ground is 16.4 + 2.7 = 19.1 m/s eastern

When the boat is heading west, its speed relative to water is -16.4m/s, its speed relative to ground is -16.4 + 2.7 = -13.7 m/s western

Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart. Calculate the force between the 2 wires. Is it attractive or repulsive? Calculate the magnetic field midway between the wires.

Answers

Answer:

$1.6* 10^(-7)$ N

$2.4* 10^(-7)$ N

Explanation:

$i_(1)$ = 1 A

$i_(2)$ = 4 A

$r$ = distance between the two wire = 5 m

$F$ = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

$F = (\mu _(o))/(4\pi )(2i_(1)i_(2))/(r)$

$F = (10^(-7))(2(1)(4))/(5)$

$F = 1.6* 10^(-7)$ N

$r'}$ = distance of each wire from the midpoint = 2.5 m

Magnetic field midway between the two wires is given as

$B = (\mu _(o))/(4\pi ) \left \left ( (2i_(2))/(r') \right - (2i_(1))/(r') \right \right ))$

$B = (10^(-7)) \left \left ( (2(4))/(2.5) \right - (2(1))/(2.5) \right \right ))$

$B = 2.4* 10^(-7)$

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down?

Answers

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 ((v_0)/(v_0-v))$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 ((v_0)/(v_0-v))$

$300 = f_0 ((343)/(343-v))$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 ((v_0)/(v_0-v/2))$

$290 = (343)((v_0)/(343-v/2))$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$((300*343) - 300v)/(290*343-145v) = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

Final answer:

The speed of the train can be determined using the Doppler effect formula.

Explanation:

The question involves the Doppler effect, which is the change in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave. In this case, the train whistle's frequency changes from 300 Hz to 290 Hz as the train approaches the station.

To find the speed of the train before and after slowing down, we can use the formula for the Doppler effect:

f' = f((v + v_o)/(v - v_s))

Where:

f' is the observed frequency
f is the source frequency
v is the speed of sound
v_o is the speed of the observer (here it is the train)
v_s is the speed of the source (here it is the speed of sound)

By substituting the given values for observed frequency (290 Hz), source frequency (300 Hz), and the speed of sound (343 m/s), we can solve for the speed of the train before and after slowing down.

Learn more about Doppler effect here:

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22. What force is necessary to accelerate a 2500kg car from rest to 20m/s over 10s?(6 Points)
2N
250N
5000N
50000N

Answers

Answer:

50000N

Explanation:

Force = mass × acceleration

= 2500 × 20

= 50000N

A T-junction combines hot and cold water streams ( = 62.4 lbm/ft3 , cp = 1.0 Btu/lbm-R). The temperatures are measured to be T1 = 50 F, T2 = 120 F at the inlets and T3 = 80 F at the exit. The pipe diameters are d1 = d3 = 2" Sch 40 and d2 = 1¼" Sch 40. If the velocity at inlet 1 is 3 ft/s what is the mass flow rate at inlet 2? (3.27 kg/s)?

Answers

Answer:

m2=3.2722lbm/s

Explanation:

Hello!

To solve this problem follow the steps below

1. Find water densities and entlapies in all states using thermodynamic tables.

note Through laboratory tests, thermodynamic tables were developed, which allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy, etc.)

through prior knowledge of two other properties, such as pressure and temperature.

D1=Density(Water;T=50;x=0)=62.41 lbm/ft^3

D2=Density(Water;T=120;x=0)=61.71 lbm/ft^3

D3=Density(Water;T=80;x=0)=62.21 lbm/ft^3

h1=Enthalpy(Water;T=50;x=0)=18.05 BTU/lbm

h2=Enthalpy(Water;T=120;x=0)=88 BTU/lbm

h3=Enthalpy(Water;T=80;x=0)=48.03 BTU/lbm

2. uses the continuity equation that states that the mass flow that enters a system is the same as the one that must exit

m1+m2=m3

3. uses the first law of thermodynamics that states that all the flow energy entering a system is the same that must come out

m1h1+m2h2=m3h3

18.05(m1)+88(m2)=48.03(m3)

divide both sides of the equation by 48.03

0.376(m1)+1.832(m2)=m3

4. Subtract the equations obtained in steps 3 and 4

m1 + m2 = m3

0.376m1 + 1.832(m2) =m3

--------------------------------------------

0.624m1-0.832m2=0

solving for m2

(0.624/0.832)m1=m2

0.75m1=m2

5. Mass flow is the product of density by velocity across the cross-sectional area

m1=(D1)(A)(v1)

internal Diameter for 2" Sch 40=2.067in=0.17225ft

$A=(\pi )/(4) D^2=(\pi )/(4) (0.17225)^2=0.0233ft^2$

m1=(62.41 lbm/ft^3)(0.0233ft^2)(3ft/S)=4.3629lbm/s

6.use the equation from step 4 to find the mass flow in 2

0.75m1=m2

0.75(4.3629)=m2

m2=3.2722lbm/s

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