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gayaneshka [121]
2 years ago
9

A circular loop with radius R= 20 cm carrying a current, I= 5 A is placed in a uniform magnetic field B = 0.2 T. (a) Find the ma

gnitude of the magnetic dipole moment of the current loop. (b) Find the maximum torque acting on the loop. (c) How much work is needed to rotate this loop in this field by 90°, starting from the initial position, 0, =0° to a final position, 0r=90° ? Hint: 0 is the angle between the magnetic dipole moment and the magnetic field​
Physics
1 answer:
Veronika [31]2 years ago
8 0
Yes it is thag answer I swear
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Consider a turnbuckle that has been tightened until the tension in wire AD is 350 N. Draw the FBD that is required to determine
Mars2501 [29]

Answer:

yes

Explanation:

yes

6 0
2 years ago
If you walk 3 kilometers in 30 minutes , what is the average speed in kilometers per hour?
Pepsi [2]

Answer:

6 km/h

Explanation:

V avg = ∆x/∆t = 3km / 30 min ×(60min/1h) = 3 km× 2 /h = 6 km/h

4 0
2 years ago
If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
sdas [7]
Data:
f_{2} = 42 Hz
n (Wave node)
V (Wave belly) 
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
f_{n} =  \frac{nV}{2L}

Wire 2 → 2º Harmonic → n = 2

f_{n} = \frac{nV}{2L}
f_{2} = \frac{2V}{2L} &#10;
2V =  f_{2} *2L
V =  \frac{ f_{2}*2L }{2}
V =  \frac{42*2L}{2}
V =  \frac{84L}{2}
V = 42L

Wire 1 → 1º Harmonic or Fundamental rope → n = 1


f_{n} = \frac{nV}{2L}
f_{1} = \frac{1V}{2L}
f_{1} =  \frac{V}{2L}

If, We have:
V = 42L
Soon:
f_{1} = \frac{V}{2L}
f_{1} = \frac{42L}{2L}
\boxed{f_{1} = 21 Hz}

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0
3 years ago
Read 2 more answers
Please help! i'm horrible at this
frez [133]

Answer:

a = Δv/t = (vf - vi)/t = (0 - 5)/4 = -1.25 m/s²

Explanation:

You may or may not need the negative sign, depending on how the question designer was thinking about the problem.

4 0
3 years ago
Una caja de 5.0kg de masa se acelera desde el reposo a través del piso mediante una fuerza a una tasa de 2.0 /s2 durante 7.0s en
Nady [450]

Responder:

<h2>490 julios </h2>

Explicación:

Se dice que el trabajo se realiza cuando una fuerza aplicada a un objeto hace que el objeto se mueva a través de una distancia. El trabajo realizado por un cuerpo se expresa mediante la fórmula;

Workdone = Fuerza * Distancia

Como Fuerza = masa * aceleración,

Workdone = masa * aceleración * distancia

Masa dada = 5.0kg, aceleración = 2.0m / s² d =?

Para obtener d, usaremos una de las leyes del movimiento,

d = ut + 1 / 2at²

u = 0 (ya que el cuerpo acelera desde el reposo) yt = 7.0s

d = 0 + 1/2 (2) (7) ²

d = 49m

Workdone = 5 * 2 * 49

Workdone = 490 Julios

4 0
3 years ago
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