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2 years ago

The electric potential at the dot in the figure is 3160 V. What is charge q?

Physics

1 answer:

PSYCHO15rus [73]2 years ago

5 0

Hi there!

Recall the equation for electric potential of a point charge:

$V = \frac{kQ}{r}$

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

$V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V$

Lower left charge's potential:

$V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V$

Add the two, and subtract from the total EP at the point:

$3160 + 1123.75 - 2247.5 = 2036.25$

The remaining charge must have a potential of 2036.25 V, so:

$2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}$

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A jogger runs by a river with a velocity of 7 m s relative to the ground. A leaf floating on the river moves with a velocity of

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Answer:

The velocity of the leaf relative to the jogger is 5 m/s.

Explanation:

Given that,

Velocity of jogger wrt to the ground, $V_j=7\ m/s$

velocity of leaf wrt the ground, $v_i=2\ m/s$

We need to find the velocity of the leaf relative to the jogger. Let it is equal to V. So, it is given by :

$V=v_j-v_i\\\\V=7-2\\\\V=5\ m/s$

So, the velocity of the leaf relative to the jogger is 5 m/s. Hence, this is the required solution.

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3 years ago

The natural force that causes you to lose power as you climb a hill is known as inertia.

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Yeah your answer is correct

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3 years ago

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

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3 years ago

The electric potential at the origin of an xy-coordinate system is 40 V. A -8.0-μC charge is brought from x = +∞ to that point.

vredina [299]

Answer:

-320 μJ.

Explanation:

Consider a point with an electrical charge of $q$ . Assume that $V$ is the electrical potential at the position of that charge. The electrical potential of that point charge will be equal to:

$\text{Potential Energy} = q \cdot V$ .

Keep in mind that since both $q$ and $V$ might not be positive, the size of the electrical potential energy might not be positive, either.

For this point charge,

$q = \rm -8.0\; \mu C$ ; (that's -8.0 microjoules, which equals to $\rm -8.0\times 10^{-6}\; J$ )
$V = \rm 40\; V$ .

Hence its electrical potential energy:

$\text{Potential Energy} = q\cdot V = \rm (-8.0\; \mu C) \times 40\; V = -320\; \mu J$ .

Why is this value negative? The electrical potential energy of a charge is equal to the work needed to bring that charge from infinitely far away all the way to its current position. Also, negative charges are attracted towards regions of high electrical potential. Bringing this $\rm -8.0\; \mu C$ negative charge to the origin will not require any external work. Instead, this process will release 320 μJ of energy. As a result, the electrical potential energy is a negative value.

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3 years ago

Ali mixes 20g of Sodium Chloride in 100g of water at 15 Degree °C. Calculate the mass of the solution

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Answer:

The mass of the solution is 120 g.

Explanation:

The mass of the solution is given by:

$m_{sol} = m_{1} + m_{2}$

Where:

$m_{sol}$ : is the mass of the solution

$m_{1}$ : is the mass of the solvent

$m_{2}$ : is the mass of the solute

In the solution, the solvent is the majority compound (in mass) and the solute is the minority (in mass), so the solvent is the water and the solute is sodium chloride.

Hence, the mass of the solution is:

$m_{sol} = m_{1} + m_{2} = 100 g + 20 g = 120 g$

I hope it helps you!

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3 years ago