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3 years ago

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When a positive charge moves in the direction of the electric field, what happens to the electrical potential energy associated

with the charge?
A. It increases.

B. It decreases.

C. It remains the same.

D. It sharply increases, and then decreases.

1 answer:

VladimirAG [237]3 years ago

6 0

Answer:

B it decreases

Explanation:

the movement of a positive test charge in the direction of an electric field would be like a mass falling downward within Earth's gravitational field. Both movements would be like going with nature and would occur without the need of work by an external force. This motion would result in the loss of potential energy

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2 years ago

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he

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Answer:8.28 km

Explanation:

Given

First it drifts $45^{\circ}$ 2.5 km

$r_1=2.5cos45 i+2.5sin45 j$

Secondly it drifts $60^{\circ}$ 4.70 km

$r_{12}=4.7cos60 i-4.7sin60 j$

After that it drifted along east direction 5.1 km

$r_{23}=5.1 i$

After that it drifts $55^{\circ}$ 7.2 km

$r_{34}=-7.2cos55 i-7.2sin55 j$

After that it drifts $5^{\circ}$ 2.8 km

$r_{54}=-2.8cos5 i+2.8sin5 j$

$r_{5O}$ = $\left [ 2.5cos45+4.7cos60+5.1-7.2cos55-2.8cos5\right ]\hat{i}$ + $\left [ 2.5sin45-4.7sin60-7.2sin55+2.8sin5\right ]\hat{j}$

$r_{5O}=2.299\hat{i}-7.95\hat{j}$

$|r_{5O}|=8.28 km$

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3 years ago

Read 2 more answers

The intensity at a distance of 6 m from a source that is radiating equally in all directions is 6.0\times 10^{-10} W/m^2. What i

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Answer:

P = 271 nW

Explanation:

If the source is radiating equally in all directions, it can be treated as a point source, so all points located at the same distance of the source, have the same intensity I, which is related to the power by the following expression:

$I = \frac{P}{A} =\frac{P}{4*\pi *r^{2} } (1)$

Solving for P, we get:

$P = I*4*\pi *r^{2} = 6.0e-10 W/m2 * 4 * \pi *(6m)^{2} =271 nW (2)$

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2 years ago

What do biologist geologist have in common how are they different

Delicious77 [7]

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2 years ago

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

a.

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

7 0

3 years ago