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3 years ago

Why is current a scalar quantity?

1 answer:

Nutka1998 [239]3 years ago

5 0

Despite current has a magnitude and a direction, like vectors, it is a scalar because it doesn't obey laws of vector addition. For instance, if we consider a junction of $90^{\circ}$ in a circuit, and two currents entering this junction, we know that the resultant current is just the algebraic sum of the two currents, not the vector sum, so it is not a vector quantity.

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If I work out rotational energy to be 102.2J which equals kg.M/s^2, and I hadn't factored time into it, would that be Joules per

Marina CMI [18]

Answer:

0.057 joules is needed to create the total rotational energy each second.

Explanation:

The energy rate is the ratio of total energy to time, which coincides with the definition of power at constant rate:

$\dot W = \frac{\Delta E}{\Delta t}$

$\dot W = \frac{102.2\,J}{\left(30\,min\right)\cdot \left(60\,\frac{s}{min} \right)}$

$\dot W = 0.057\,\frac{J}{s}$

$\dot W = 0.057\,W$

0.057 joules is needed to create the total rotational energy each second.

7 0

3 years ago

Which of the following molecules is correctly paired with its macromolecule class? (2 points)

Mazyrski [523]

Answer:

adjust chosen Karl Karl jam halted Bernie

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3 years ago

Show that the energy of a magnetic dipole m in a magnetic field B is U--m B

Juli2301 [7.4K]

Answer:

showm

Explanation:

Consider a dipole having magnetic moment 'm' is placed in magnetic field $\vec{B}$ then the torque exerted by the field on the dipole is

$\tau = m\times B$

$\tau=mBsin\alpha$

Now to rotate the dipole in the field to its final position the work required to be done is

$U=\int \tau d\alpha$

$U=\int mBsin\alpha d\alpha$

$U= -mBcos\alpha$

$U=-\vec{m\times \vec{B}}$

Minimum energy mB is for the case when m is anti parallel to B.

Minimum energy -mB is for the case when m is parallel to B.

5 0

3 years ago

The drawing shows a plot of the output emf of a generator as a function of time t. The coil of this device has a cross-sectional

adell [148]

This question is incomplete, the missing image uploaded along this answer below;

Answer:

a) frequency is 2.381 Hz

b) the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field is 0.5199 T

Explanation:

Given that;

cross sectional Area A = 0.018 m²

Number of turns N = 200

from the diagram. maximum time T = 0.42 sec

a) the frequency f of the generator in hertz

frequency = 1 / T

we substitute

frequency = 1 / 0.42 = 2.381 Hz

Therefore, frequency is 2.381 Hz

b) the angular frequency in rad/s

angular speed ω = 2πf

we substitute

ω = 2π × 2.381

ω = 14.96 rad/s

Therefore, the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field.

to determine the magnitude of the magnetic field, we use the following expression;

e = NBAω

from the diagram, e = 28.0 V

so we substitute

28.0 V = 200 × B × 0.018 × 14.96

28 = 53.856B

B = 28 / 53.856

B = 0.5199 T

Therefore, the magnitude of the magnetic field is 0.5199 T

6 0

3 years ago

What explains the information needed to calculate speed and velocity?

Ede4ka [16]

To calculate speed, you need to know the total distance covered
and the time it took to cover the distance.

To calculate velocity, you need the starting location, the ending location,
and the time it took to go from start to finish.

3 0

3 years ago