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Elodia [21]
3 years ago
8

The potential difference between two points is 100 V. If a particle with a charge of 2 C is transported from one of these points

to the other, what is the magnitude of the work done?
Physics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0

1 volt = 1 joule/coulomb

100 volts = 100 joules/coulomb

Work = (100 J/C) x (2 C)

Work = (+ or -) 200 J

The work could be positive or negative. It depends on direction of the potential difference, direction the charge is moved, sign of the charge, and whether we mean work done by the mover or by the potential difference.

But the MAGNITUDE of the work is 200J .

Aleonysh [2.5K]3 years ago
4 0

Answer: 200 J

Explanation: In order to explain this we have consider that the work done in a electric field is given by:

Work= Q*ΔV=2*100=200J

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A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo
BARSIC [14]

We are given that,

\frac{dx}{dt} = 4ft/s

We need to find \frac{d\theta}{dt} when x=8ft

The equation that relates x and \theta can be written as,

\frac{x}{6} tan\theta

x = 6tan\theta

Differentiating each side with respect to t, we get,

\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}

\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}

\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}

Replacing the value of the velocity

\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2

\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta

The value of cos \theta could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

cos\theta = \frac{6}{10}

\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2

\frac{d\theta}{dt} = \frac{24}{25}

Search light is rotating at a rate of 0.96rad/s

4 0
2 years ago
Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520c as measur
Over [174]

Answer:

the magnitude of the velocity of one particle relative to the other is 0.9988c

Explanation:

Given the data in the question;

Velocities of the two particles = 0.9520c

Using Lorentz transformation

Let relative velocity be W, so

v_r = ( u + v ) / ( 1 + ( uv / c²) )

since each particle travels with the same speed,

u = v

so

v_r = ( u + u ) / ( 1 + ( u×u / c²) )  

v_r = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )  

we substitute

v_r = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )  

v_r = 1.904c / ( 1 + 0.906304 )

v_r = 1.904c / 1.906304

v_r = 0.9988c

Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c

5 0
3 years ago
Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

b) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{743}{2\pi}

          f = 118.25 Hz

c) wavelength

        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

        \lambda= \dfrac{2\pi}{7.02}

        \lambda= 0.895\ m

d) speed

         v = \dfrac{\omega}{k}

         v = \dfrac{743}{7.02}

                v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

  T = \dfrac{v^2\ m}{L}

  T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}

      T = 27.87 N

g) Power transmitted by the wave

  P = \dfrac{1}{2}m\ v \omega^2\ A^2

  P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2

      P = 0.438 W

5 0
3 years ago
Which things determine an element's electronegativity? Check all that apply. A. The size of the atom of that element B. Whether
KengaRu [80]
D. The number of electrons
6 0
3 years ago
Please help! Will give brainly, 50 points!! I'm stuck with this question and I don't get it!!
IrinaVladis [17]

Answer:

3.52176 x 10^-10 N

Explanation:

Fg = 3.52176 x 10^-10 Newton

8 0
3 years ago
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