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Kobotan [32]
3 years ago
12

A round loop of wire carries a current of 100 A, has a radius of 10 cm, and its normal (vector) makes an angle of 30∘ with a mag

netic field of 0.524 T. 1) What is the torque on the loop? (Express your answer to two significant figures.)
Physics
1 answer:
Oliga [24]3 years ago
7 0

Answer:

0.823 Nm

Explanation:

current, i = 100 A

radius, r = 10 cm

Angle between the normal and the magnetic field, θ = 30°

Magnetic field, B = 0.524 T

Torque is defined as the

\tau = i \times A\ B\times sin\theta

\tau = 100 \times 3.14\times 0.1\times 0.1\ 0.524\times 0.5

Torque = 0.823 Nm

Thus, the torque is 0.823 Nm.

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Formulating a Hypothesis: Part I Since the investigative question has two variables, you need to focus on each one separately. T
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Answer : The kinetic energy depends directly on the mass of a particle.

Explanation :

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KE=\dfrac{1}{2}mv^2

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v is the velocity with which it is moving

Kinetic energy is due to the motion of the particle.

So, the kinetic energy of a particle is directly proportional to its mass.

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A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ
DochEvi [55]

Answer:

a)  k=19.6N/m

b)  V_m=0.81m/s

c)  a_m=6.561m/s^2

d)  K.E=0.096J

e)  T=0.78sec &F=1.29sec

f)   mx'' + kx' =0

Explanation:

From the question we are told that:

Stretch Length L=0.150m

Mass m=0.30kg

Total stretch lengthL_t=0.150+0.100=>0.25

a)

Generally the equation for Force F on the spring is mathematically given by

F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}

k=19.6N/m

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

V_m=A\omega

Where

A=Amplitude

A=0.100m

And

\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s

Therefore

V_m=A\omega\\\\V_m=8.1*0.1

V_m=0.81m/s

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

a_m=\omega^2A

a_m=8.1^2*0.1

a_m=6.561m/s^2

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}*0.3*0.8^2

K.E=0.096J

e)

Generally the equation for  the period T is mathematically given by

\omega=\frac{2\pi}{T}

T=\frac{2*3.142}{8.1}

T=0.78sec

Generally the equation for  the Frequency is mathematically given by

F=\frac{1}{T}

F=1.29sec

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

mx'' + kx' =0

Where

'= signify order of differentiation

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Determine the equivalent capacitance between points a and <br> b.
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