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Juliette [100K]
3 years ago
7

What is the strength of the electric field in a region where the electric potential is constant?

Physics
1 answer:
Roman55 [17]3 years ago
4 0

Answer:

Where the electric potential is constant, the strength of the electric field is zero.

Explanation:

As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e

Eₓ = - dV / dx        ----------(i)

From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.

<em>Therefore, a constant electric potential means that electric field is zero.</em>

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The fundamental mode of vibration of a wave is defined as that which has the ___________ frequency
kolbaska11 [484]

The fundamental mode of vibration of a wave is defined as that which has the LOWEST frequency  

5 0
3 years ago
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The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune
Veronika [31]
Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
\frac{r^3}{T^2}= \frac{GM}{4 \pi^2} (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
1 AU = 1.5 \cdot 10^{11} m
so the radius of the orbit is
r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
T=\sqrt{ \frac{4 \pi^2}{GM} r^3} =  \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years
3 0
3 years ago
A 55 kgkg meteorite buries itself 5.5 mm into soft mud. The force between the meteorite and the mud is given by F(x)F(x) = (630
sashaice [31]

Answer:

W = 1.44 10⁻⁷ J

Explanation:

The expression for the job is

          W = ∫ F. dx

Where the point is the scalar product in this case the direction of the meteor and the depth is parallel, whereby the scalar product is reduced to the ordinary product

          W = 630 ∫ x³ dx

          W = 630 x⁴ / 4

Let's evaluate between the lower limit x = 0, w = 0 to the upper limite the point at x = 5.5 10⁻³ m

            W = 157.5 ((5.5 10⁻³)⁴ -0)

            W = 1.44 10⁻⁷ J

6 0
3 years ago
A car travels 40 kilometers at an average speed of 80 km/h and then travels 40 kilometers at
miv72 [106K]

Answer:53.3m/s

Explanation:

time for first travel

time=distance/speed

time=40/80

time=0.5seconds

time for second travel:

time=distance/speed

time=40/40

time=1seconds

total time=0.5+1

total time=1.5seconds

Then average speed for 80km trip:

Speed=distance/time

Speed=80/1.5

Speed=53.3m/s

3 0
4 years ago
A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w
frosja888 [35]

Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) n_{2} = 1.349

(c) v_{1} = 2.04\times 10^{8}\ m/s

(d) v_{2} = 2.22\times 10^{8}\ m/s

Solution:

As per the question:

Refractive index of medium 1, n_{1} = 1.47

Angle of refraction for medium 1, \theta_{1} = 59^{\circ}

Angle of refraction for medium 2, \theta_{1} = 69^{\circ}

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

n_{1}sin\theta_{1} = n_{2}sin\theta_{2}

where

n_{2} = Refractive Index of medium 2

Now,

n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}

n_{2} = 1.349

(c) To calculate the velocity of light in medium 1:

We know that:

Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}

Thus for medium 1

n_{1} = \frac{c}{v_{1}

v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s

(d) To calculate the velocity of light in medium 2:

For medium 2:

n_{2} = \frac{c}{v_{2}

v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s

5 0
3 years ago
Read 2 more answers
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