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andriy [413]
3 years ago
7

What is electrical energy​

Physics
1 answer:
creativ13 [48]3 years ago
8 0

Explanation:

it is energy that flows of electric charge it has the ability to do work or apply force to move an object.

You might be interested in
Question 27 a motorboat takes 3 hours to travel 144km going upstream. the return trip takes 2 hours going downstream. what is th
Vedmedyk [2.9K]
In this item, we let x be the rate of the boat in still water and y be the rate of the current.

Upstream. When the boat is going upstream, the speed in still water is deducted by the speed of the current because the boat goes against the water. The distance covered is calculated by multiplying the number of hours and the speed.
                 (x - y)(3) = 144

Downstream. The speed of the boat going downstream is equal to x + y because the boat goes with the current.
                (x + y)(2) = 144

The system of linear equations we can use to solve for x is,
                  3x - 3y  = 144
                  2x + 2y = 144

We use either elimination or substitution.
 
We solve for the y of the first equation in terms of x,
                   y =  -(144 - 3x)/3

Substitute this to the second equation,
            2x + 2(-1)(144 - 3x)/3 = 144
The value of x from the equation is 60

<em>ANSWER: 60 km/h</em>
5 0
3 years ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Veseljchak [2.6K]

Answer:

The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})

0.00075(400-v_{f}^2)=0.18&#10;

400-v_{f}^2=\dfrac{0.18}{0.00075}

-v_{f}^2=240-400

v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

7 0
2 years ago
Which layer of the earth can be seen from the open hole
umka21 [38]

Answer:

the inner core

Explanation:

6 0
3 years ago
Read 2 more answers
You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli
FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

f(x,t)=Acos(kx-\omega t)

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}

k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}

T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m

b) Hence, the wave function is:

f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)

c) for x=3m you have:

f(3,t)=0.075cos(1.047*3-12.56t)

d) the speed of the medium:

\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)

you can see the velocity of the medium for example for x = 0:

v=\frac{df}{dt}=13.15cos(12.56t)

7 0
3 years ago
A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.
Fiesta28 [93]

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

d=30,000 pc =9.26\cdot 10^{20} m

The speed of the cosmic ray is

v=0.98 c

where

c=3.0 \cdot 10^8 m/s is the speed of light. Substituting,

v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s

Converting into years,

T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}

And substituting v = 0.98c, we find:

T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years

3 0
3 years ago
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