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2 years ago

The electric field 30cm from a van de Graaff generator is measured to be 28,300N/C. What is the charge of the van de Graaf?

Physics

1 answer:

gizmo_the_mogwai [7]2 years ago

8 0

Answer:

Explanation:

EWAN KO LANG DIN BASTA YAN ALAM KO

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Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

4 0

3 years ago

Select the correct answer from each drop-down menu.

earnstyle [38]

Answer:

a transverse (sort of a plot of a sine or cosine graph, basically)

b longitudinal

c Electromagnetic (an electric wave and a magnetic wave travelling together at right angles to each other)

Explanation:

7 0

3 years ago

What does it mean when a wave’s amplitude increases?

sashaice [31]

Answer:

the wave is carrying more energy

Explanation:

trust me broski

7 0

2 years ago

please help In a video game, a ball moving at 0.6 meter/second collides with a wall. After the collision, the velocity of the ba

viva [34]

Answer:

the acceleration during the collision is: - 5 $\frac{m}{s^2}$

Explanation:

Using the formula:

$a=\frac{\Delta\,v}{\Delta\,t}$

we get:

$a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}$

4 0

3 years ago

The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. a typical axon carries a peak cu

Gemiola [76]

Answer:

$6.66\cdot 10^{-12}T$

Explanation:

The magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

I is the current

r is the distance from the wire

In this problem we have

$I=0.040 \mu A=4\cdot 10^{-8}A$

r = 1.2 mm = 0.0012 m

So the magnetic field strength is

$B=\frac{(4\pi \cdot 10^{-7} H/m)(4\cdot 10^{-8}A)}{2\pi (0.0012 m)}=6.66\cdot 10^{-12}T$

5 0

3 years ago