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Mila [183]
3 years ago
9

A charge per unit length given by l(x) = bx, where b = 12 nc/m2, is distributed along the x axis from x = +9.0 cm to x = +16 cm.

if the electric potential at infinity is taken to be zero, what is the electric potential at the point p on the y axis at y = 12 cm?

Physics
1 answer:
Pavel [41]3 years ago
4 0
Hopefully this will help you.

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A flag is waved 3.2 m above the surface of a flat pool of water. When viewed from under the water, what is the magnification of
jek_recluse [69]

Answer:

1.33

Explanation:

For an optical instrument, the magnification ratio of the apparent diameter of the image to that of the object.

Mathematically, from the given information;

Magnification= \dfrac{n_{water}}{n_{air}}

where;

n_{water} =1.33\\ \\ n_{air} = 1.00

= \dfrac{1.33}{1.00} \\ \\=  \mathbf{1.33}

5 0
3 years ago
Q 28.7: A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 T. What is the maximum Hall
Zinaida [17]

Answer:the  maximum Hall voltage across the strip= 0.00168 V.

Explanation:

The Hall Voltage is calculated using

Vh= B x v x  w

Where

B is the magnitude of the magnetic field, 5.6 T

v is the speed/  velocity of the strip, = 25 cm/s  to m/s becomes 25/100=0.25m/s

and w is the width of the strip=  1.2 mm to meters becomes 1.2 mm /1000= 0.0012m

 Solving

Vh= 5.6T x 0.25m/s x 0.0012m

=0.00168T.m²/s

=0.00168Wb/s

=0.00168V

Therefore, the  maximum Hall voltage across the strip=0.00168V

3 0
3 years ago
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
1.5 x 10^3 standard notation
poizon [28]

Answer:

1500

Explanation:

3 0
3 years ago
Suppose a standing wave created on a spring that is 10.5 m long has a speed of 14 m/s and a frequency of 2 Hz. How many nodes an
dimulka [17.4K]

Answer:

If the two waves have the same amplitude and wavelength, then they alternate between ... In fact, the waves are in phase at any integer multiple of half of a period: ... The propagation velocity of the waves is 175 m/s.

Explanation:

please ask me in brainlist ok

6 0
3 years ago
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