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KIM [24]
2 years ago
6

A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0

) to the point (x2, y2). Please use k for Coulomb's constant rather than writing it out as (1/4πϵ0).
How much work W is done by the electrostatic force on the moving point charge?
Physics
1 answer:
Scilla [17]2 years ago
3 0

Answer:

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Explanation:

Position of charge q₁ is (0,0)

Position of charge q₂ is (x₁,0)

So, the electric potential energy between the charges is given by :

U_1=k\dfrac{q_1q_2}{x_1}

Now the position of charge q₂ has been changes from (x₁,0) to (x₂,y₂). Now, electric potential energy between the charges is :

U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}

We know form the work energy theorem that, the change in potential energy is equal to the work done. Mathematically, it is given by :

W=-\Delta U

W=-(U_2-U_1)

W=(U_1-U_2)

W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Hence, the work done by the electrostatic force on the moving point charge is kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}}). Hence, this is the required solution.

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Sea breezes that occur near the shore are attributed to a difference between land and water with respect to what property?
ddd [48]

Answer:

a. mass density

Explanation:

<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>

  • When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.

<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>

7 0
2 years ago
It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 110 m above a raging riv
zzz [600]

Answer:

h=20.66m

Explanation:

First we need the speed when the cord starts stretching:

V_2^2=V_o^2-2*g*\Delta h

V_2^2=-2*10*(-31)

V_2=24.9m/s   This will be our initial speed for a balance of energy.

By conservation of energy:

m*g*h+1/2*K*(h_o-l_o-h)^2-m*g*(h_o-l_o)-1/2*m*V_2^2=0

Where

h is your height at its maximum elongation

h_o is the height of the bridge

l_o is the length of the unstretched bungee cord

800h+21*(79-h)^2-63200-24800.4=0

21h^2-2518h+43060.6=0 Solving for h:

h_1=20.66m  and h_2=99.24m  Since 99m is higher than the initial height of 79m, we discard that value.

So, the final height above water is 20.66m

6 0
2 years ago
Read 2 more answers
A 40-mH ideal inductor is connected in series with a 50 Ω resistor through an ideal 15-V DC power supply and an open switch. If
sergey [27]

Answer:i=300 mA

Explanation:

Given

inductance(L)=40 mH

Resistor(R)=50 \Omega

Voltage(V)=15 V

Time constant(\tau)=\frac{L}{R}

\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}

current i_0=\frac{V}{R}

i_0=\frac{15}{50}=0.3 A

Current as a function of time is given by

i=i_0\left ( 1-e^{-\frac{t}{\tau }}\right )

i=0.3\times 0.9998

i= 299.95 mA

6 0
2 years ago
*An inductor is capable of dissipating 50W of heat energy when a current 0.8A flows through it at a certain frequency. Calculate
ale4655 [162]
I think that your answer would be D
8 0
2 years ago
A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position. Why?
spin [16.1K]

Answer:

The body's rotational inertia is greater in layout position than in tucked position. Because the body remains airborne for roughly the same time interval in either position, the gymnast must have much greater kinetic energy in layout position to complete the backflip.

Explanation:

A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

When the body is straight , its moment of rotational inertia is more than the case when he folds his body round. Hence rotational inertia ( moment of inertia x angular velocity ) is also greater. To achieve that inertia , there is need of greater imput of energy in the form of kinetic energy  which requires greater effort.

So a gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

6 0
2 years ago
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