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igor_vitrenko [27]

2 years ago

13

Define electric potential energy

2 answers:

Inessa05 [86]2 years ago

6 0

Electric potential energy, or Electrostatic potential energy, is a potential energy (measured in joules) that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system. An object may have electric potential energy by virtue of two key elements: its own electric charge and its relative position to other electrically charged objects.

Verizon [17]2 years ago

3 0

Electric potential energy, or Electrostatic potential energy, is a potential energy that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system.

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Because atoms of elements in the same group of the periodic table have the same number of neutrons, they have similar properties

AURORKA [14]

False. What actually determines the properties of elements are the electrons, or aka valence electrons. They are used to bond, which determines its properties.

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2 years ago

Name 3 elements in group one

Vinil7 [7]

Answer:

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Explanation:

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2 years ago

DETERMINE THE BCD EQUIVALENT OF 1100111110101001

Solnce55 [7]

Answer: It states that the BCD equivalent would be 0001000100000000000100010001000100010000000100000001000000000001.

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2 years ago

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

2)

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

3)

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

4)

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

7 0

2 years ago

A 1500 kg truck travelling north at 60 km/hr collides with a 1200 kg car moving east at 15km/hr. If the two cars remain locked t

Alenkinab [10]

Answer:

33.33j+6.67i km/hr

Explanation:

From the law of conservation of momentum,

Applying,

mu+m'u' = V(m+m')............... Equation 1

Where m = mass of the truck, m' = mass of the car, u = initial velocity of the truck, u' = initial velocity of the car, V = Final velocity.

Note: let j represent the north, and i represent the east

From the question,

Given: m = 1500 kg, u = 60j, m' = 1200 kg, u' = 15i

Substitute these values into equation 1

1500*60j+1200*15i = V(1500+1200)

90000j+18000i = 2700V

V = (90000j+18000i)/2700

V = 33.33j+6.67i km/hr

6 0

2 years ago