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Aneli [31]
3 years ago
10

Do magnetic field lines ever start or end anywhere?

Physics
1 answer:
nevsk [136]3 years ago
4 0

Answer:

No

Explanation:

The lines of the field of a magnet don't begin or stop at anyplace, they generally make shut circles or loops and will proceed inside magnet (however here and there they are not drawn along these lines). We require an approach to show the bearing of the field.  

The field lines of a magnet don't simply end at the magnetic tip. They go directly through it, so that inside the magnet the magnetic field lines indicates from the south to the north pole.

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un motor electric efectueaza un lucru mecanic de 864 j in 0.5 min.tensiunea la bornele sale este de 12v si este parcurs de un cu
GrogVix [38]

Answer:

96%

Explanation:

To find the values of the motor efficiency you use the following formula:

E=\frac{P_o}{P_i}100

P_o: output power = 864J/0.5min=864J/30s=28.8W

P_i: input power = I*V = (3A)(12V) = 36W

By replacing this values you obtain:

E=\frac{28.8W}{30W}*100=96\%

hence, the motor efficiency is about 96%

traslation:

Pentru a găsi valorile eficienței motorului, utilizați următoarea formulă:

P_o: putere de ieșire = 864J / 0.5min = 864J / 30s = 28.8W

P_i: putere de intrare = I * V = (3A) (12V) = 36W

Înlocuind aceste valori obțineți:

prin urmare, eficiența motorului este de aproximativ 96%

3 0
3 years ago
A person on a daily diet of 2,500 calories should get no_more than <br> calories from fat each day.
ivann1987 [24]

Answer:

35%

Explanation:

3 0
3 years ago
Read 2 more answers
The first law of thermodynamics relates the heat transfer into or out of a system to the change of internal and the work done on
Artyom0805 [142]

Answer:

The heat transferred into the system is 183.5 J.

Explanation:

The first law of thermodynamics relates the heat transfer into or out of a system to the change of internal and the work done on the system, through the following equations.

ΔU = Q - W

where;

ΔU  is the change in internal energy

Q is the heat transfer into the system

W is the work done by the system

Given;

ΔU = 155 J

W = 28.5 J

Q = ?

155 = Q - 28.5

Q = 155 + 28.5

Q = 183.5 J

Therefore, the heat transferred into the system is 183.5 J.

4 0
2 years ago
If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular
erastova [34]

Answer;

D. The car would begin to move in the direction it was headed in a straight line.

Explanation;

-Centripetal force is any net force causing uniform circular motion. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.

-The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

-Therefore,If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced then the car would begin to move in the direction it was headed in a straight line.

6 0
3 years ago
Read 2 more answers
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

3 0
3 years ago
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