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maxonik [38]
2 years ago
9

If the electric potential in a region is given by v(x)=7/x2 the x component of the electric field in that region is

Physics
1 answer:
lions [1.4K]2 years ago
6 0

Hi there!

Recall that:
\Delta V = -\int\limits^a_b {E \cdot } \, dx

Given an electric field, the potential difference can be solved by using integration. Similarly:
E = -\frac{dV}{dx}

We can differentiate the electric potential equation to solve for the electric field.

Use the power rule:
\frac{dy}{dx} x^n = nx^{n - 1}

Differentiate the given equation.

-\frac{dV}{dx}\frac{7}{x^2} =- \frac{dV}{dx}7x^{-2} = -(-14x^{-3}) = \frac{14}{x^3}

Or:
\boxed{E(x) = \frac{14}{x^3}}

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A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught
meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2} (Eq. 1)

Where:

y_{o} - Initial height of the softball, measured in meters.

y - Final height of the softball, measured in meters.

v_{o} - Initial velocity of the softball, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

If we know that y = y_{o}, t = 3.56\,s and g = -9.807\,\frac{m}{s^{2}}, the initial velocity of the softball is:

v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0

3\cdot v_{o} -44.132\,m= 0

v_{o} = 14.711\,\frac{m}{s}

The initial velocity of the softball is 14.711 meters per second.

8 0
3 years ago
A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring
Dennis_Churaev [7]

Answer: 6067.5 N

Explanation:

Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.

Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.

5 0
2 years ago
Wave C has an amplitude of 1 and wave D has an amplitude of 3 as shown below. What will happen when the trough of wave C meets t
gtnhenbr [62]
They will interfere to create a crest with an amplitude of 2
4 0
3 years ago
Read 2 more answers
30.42
CaHeK987 [17]

Answer: Try C

Explanation:

It's the only one that makes since.

7 0
2 years ago
A:10i - 2j -4k and B: i +7j - k. Determine |A-B| ​
maksim [4K]

<em>A</em> - <em>B</em> = (10<em>i</em> - 2<em>j</em> - 4<em>k</em>) - (<em>i</em> + 7<em>j</em> - <em>k</em>)

<em>A</em> - <em>B</em> = 9<em>i</em> - 9<em>j</em> - 3<em>k</em>

|<em>A</em> - <em>B</em>| = √(9² + (-9)² + (-3)²) = √189 = 3√19

8 0
2 years ago
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