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GREYUIT [131]
4 years ago
13

How much work does the electric field do in moving a proton from a point with a potential of +155 v to a point where it is -65 v

? express your answer in joules?
Physics
1 answer:
Zigmanuir [339]4 years ago
7 0
The work done by the electric field is equal to the loss of electric potential energy of the proton in moving from its initial location to its final location:
W=-\Delta U = -q \Delta V = -q (V_f -V_i)
where q=1.6 \cdot 10^{-19}C is the proton charge, V_f = -65 V and V_i=+155 V are the voltages in the final and initial locations. Substituting, we get
W=-(1.6 \cdot 10^{-19}C)(-65 V-(+155 V))=3.5 \cdot 10^{-17}J
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3. an observer is at the midpoint between two coherent in-phase sources of sound of frequency 780 hz. what is the minimum distan
baherus [9]

Answer:

Explanation:

wavelength of sound = velocity / frequency

= 340 / 780

= .4359 m

Let the observer be at equal distance d from in phase source .

let it moves by distance x for destructive interference .

path difference from source

= d + x - (d - x )

=  2x

for destructive interference

path difference = wave length / 2

2x = .4359 / 2 m

x = .4359 / 4

= .108975 m

10.9 cm

so observer must move by distance 10.9 cm towards on of the centers.

6 0
3 years ago
An object moves with constant acceleration 3.45 m/s2 and over a time interval reaches a final velocity of 14.0 m/s.
Sav [38]
Assuming the object was originally at rest, it must have been traveling for
14.0/3.45 = 4.06 seconds
4 0
3 years ago
Which of the following is an example of a buoyant force acting on an object?
baherus [9]

Answer:

  1. option c theobject floats on top of a fluid
3 0
3 years ago
Each shot of the laser gun favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the dis
Alisiya [41]

Answer:

a) 4.94e9 J  b) 1.07e10 J

Explanation:

The electric potential energy stored in a capacitor, expressed in terms of the value of the capacitance C, and the voltage between its terminals V, is as follows:

U =\frac{1}{2}*C*V^{2}

a) For the original capacitor, we can find directly U as follows:

U =\frac{1}{2}*1.81F*(73.9e3)^{2} V2 = 4.94e9 J  

U = 4.94*10⁹ J

b) Prior to find the electric potential energy of the upgraded capacitor, we need to find out the value of the capacitance C of this capacitor, which is identical to the original, except that has a different dielectric constant.

As the capacitance is proportional to the dielectric constant, we can write the following proportion:

ε₂ / ε₁ = \frac{943}{435}= \frac{C2}{C1} =\frac{Cx}{1.81F}

Cx =\frac{1.81F*943}{435} = 3.92 F

Once calculated the new value of the capacitance, as V remains the same, we can find the electric potential energy for the upgraded capacitor as follows:

U =\frac{1}{2}*3.92F*(73.9e3)^{2} V2 = 1.07e10 J

⇒ U = 1.07*10¹⁰ J

8 0
3 years ago
Am i correct? If not then which one
cupoosta [38]

Answer:

Yes, it's correct

Explanation:

Newton's second Law states that the acceleration of an object is proportional to the net force applied on it, according to the equation:

F=ma

where

F is the net force on the object

m is the mass of the object

a is the acceleration of the object

We can re-arrange the previous equation in order to solve explicitely for a, the acceleration, and we find:

F=ma\\\frac{F}{m}=\frac{ma}{m}\\\frac{F}{m}=a\\a=\frac{F}{m}

So, we see that the acceleration is proportional to the net force and inversely proportional to the mass of the object.

4 0
4 years ago
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