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just olya [345]

3 years ago

6

A bead with a mass of 0.050 g and a charge of 20 nC is free to slide on a vertical rod. At the base of the rod is a fixed 20 nC

charge. In equilibrium, at what height above the fixed charge does the bead rest?

1 answer:

erma4kov [3.2K]3 years ago

3 0

Answer:

height above the fixed charge is 8.571 cm

Explanation:

given data

mass m = 0.050 g = 5 × $10^{-5}$ kg

charge bead q1 = 20 nC = 20 × $10^{-9}$ C

charge base q2 = 20 nC = 20× $10^{-9}$ C

to find out

what height above the fixed charge does the bead rest

solution

we know that when charge at rest then downward gravitational force is balance by electrostatic force so

$mg = k\frac{q1q2}{r^2}$ .............1

here k is 9 × $10^{9}$ Nm²/C² and g = 9.8 m/s² and r is height of bread

put here all value in equation 1

$5*10^{-5}*9.8 = 9*10^{9} \frac{20*10^{-9}*20*10^{-9}}{(r^2}$

r² = 7.3469 × $10^{-3}$

r = 0.08571 m = 8.571 cm

so height above the fixed charge is 8.571 cm

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Answer:

15.4 kg.

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

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

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.

Given: m = 7.7 kg, u' = 0 m/s (at rest)

Let: u = x m/s, and V = 1/3x m/s

Substitute into equation 1

7.7(x)+m'(0) = 1/3x(7.7+m')

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7 0

3 years ago

Read 2 more answers

What is black body radiation? Explain in detail.

tangare [24]

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3 0

2 years ago

SashulF [63]

Answer:

a. Final velocity, V = 2.179 m/s.

b. Final velocity, V = 7.071 m/s.

Explanation:

<u>Given the following data;</u>

Acceleration = 0.500m/s²

a. To find the velocity of the boat after it has traveled 4.75 m

Since it started from rest, initial velocity is equal to 0m/s.

Now, we would use the third equation of motion to find the final velocity.

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.

Substituting into the equation, we have;

$V^{2} = 0^{2} + 2*0.500*4.75$

$V^{2} = 4.75$

Taking the square root, we have;

$V^{2} = \sqrt {4.75}$

<em>Final velocity, V = 2.179 m/s.</em>

b. To find the velocity if the boat has traveled 50 m.

$V^{2} = 0^{2} + 2*0.500*50$

$V^{2} = 50$

Taking the square root, we have;

$V^{2} = \sqrt {50}$

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8 0

2 years ago

Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw

Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

$\displaystyle{U=\frac{q}{4\pi \epsilon_0r}}$ (1)

where $q$ is the charge of the particle, $\epsilon_0$ the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and $r$ is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

$\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}$

Substituting the values $q=1.602 \cdot10^{-19}\ C$ , $\displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}}$ and $r=0.5 \cdot 10^{-9} m$ we obtain:

$\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}$

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

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3 years ago

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Answer:

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5 0

3 years ago