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Nadusha1986 [10]

2 years ago

10

An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect

ron reaches a closest distance of 15 cm before being repelled.

1 answer:

melisa1 [442]2 years ago

3 0

Answer:

The plate's surface charge density is $-8.056\times10^{-9}\ C/m^2$

Explanation:

Given that,

Speed = 9800 km/s

Distance d= 75 cm

Distance d' =15 cm

Suppose we determine the plate's surface charge density?

We need to calculate the surface charge density

Using work energy theorem

$W=\Delta K.E$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2$

Here, final velocity is zero

$W=0-\dfrac{1}{2}mv_{i}^2$ ...(I)

We know that,

$W=-Fd$

$W=-E\times e\times d$

$W=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$ ...(II)

From equation (I) and (II)

$-\dfrac{1}{2}mv_{i}^2=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$

Charge is negative for electron

$\lambda=\dfrac{mv^2\epsilon_{0}}{(-e)d}$

Put the value into the formula

$\lambda=-\dfrac{9.1\times10^{-31}\times(9800\times10^{3})^2\times8.85\times10^{-12}}{1.6\times10^{-19}\times(75-15)\times10^{-2}}$

$\lambda=-8.056\times10^{-9}\ C/m^2$

Hence, The plate's surface charge density is $-8.056\times10^{-9}\ C/m^2$

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In a distant solar system, a giant planet has

sergeinik [125]

Answer:

mass of the planet: $5.9\,10^{26}\,kg$

Explanation:

When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):

$m\frac{v^2}{R}=G\frac{M\,m}{R^2}$

where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:

$m\frac{v^2}{R}=G\frac{M\,m}{R^2}\\v^2=G\frac{M}{R}$

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.

We know the orbital radius R ( $5.32\,10^5\,km=5.32\,10^8\,m$ , the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.

We know that the moon makes a full circumference ( $2\,\pi\,R$ ) in 388800 seconds, therefore its tangential velocity is:

$v=\frac{2\,\pi\,5.32\,10^8}{388800} \frac{m}{s} \\v=8.6\,10^3\,\frac{m}{s}$

where we rounded the velocity to one decimal.

Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.

Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

$v^2=G\frac{M}{R}\\M=\frac{v^2\,R}{G} \\M=\frac{(8.6\,10^3)^2\,5.32\,10^8}{6.67\,10^{-11}}kg\\M=5.9\,10^{26}\,kg$

8 0

2 years ago

A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows:

Nikitich [7]

Answer:

r ≥ R, E = Q / (4πR²ε₀)

r ≤ R, E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

Maximum at r = ⅔ R

Maximum field of E = Q / (3πε₀R²)

Explanation:

Gauss's law states:

∮E·dA = Q/ε₀

What that means is, if you have electric field vectors E passing through areas dA, the sum of those E vector components perpendicular to the dA areas is equal to the total charge Q divided by the permittivity of space, ε₀.

a) r ≥ R

Here, we're looking at the charge contained by the entire sphere. The surface area of the sphere is 4πR², and the charge it contains is Q. Therefore:

E(4πR²) = Q/ε₀

E = Q / (4πR²ε₀)

b) r ≤ R

This time, we're looking at the charge contained by part of the sphere.

Imagine the sphere is actually an infinite number of shells, like Russian nesting dolls. For any shell of radius r, the charge it contains is:

dq = ρ dV

dq = ρ (4πr²) dr

The total charge contained by the shells from 0 to r is:

q = ∫ dq

q = ∫₀ʳ ρ (4πr²) dr

q = ∫₀ʳ ρ₀ (1 − r/R) (4πr²) dr

q = 4πρ₀ ∫₀ʳ (1 − r/R) (r²) dr

q = 4πρ₀ ∫₀ʳ (r² − r³/R) dr

q = 4πρ₀ (⅓ r³ − ¼ r⁴/R) |₀ʳ

q = 4πρ₀ (⅓ r³ − ¼ r⁴/R)

Since ρ₀ = 3Q/(πR³):

q = 4π (3Q/(πR³)) (⅓ r³ − ¼ r⁴/R)

q = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴)

Therefore:

E(4πr²) = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / ε₀

E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

When E is a maximum, dE/dr is 0.

First, simplify E:

E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

E = Q (4 (r³/R³) − 3 (r⁴/R⁴)) / (4πr²ε₀)

E = Q (4 (r/R³) − 3 (r²/R⁴)) / (4πε₀)

Take derivative and set to 0:

dE/dr = Q (4/R³ − 6r/R⁴) / (4πε₀)

0 = Q (4/R³ − 6r/R⁴) / (4πε₀)

0 = 4/R³ − 6r/R⁴

0 = 4R − 6r

r = ⅔R

Evaluating E at r = ⅔R:

E = Q (4 (⅔R / R³) − 3 (⁴/₉R² / R⁴)) / (4πε₀)

E = Q (8 / (3R²) − 4 / (3R²)) / (4πε₀)

E = Q (4 / (3R²)) / (4πε₀)

E = Q (1 / (3R²)) / (πε₀)

E = Q / (3πε₀R²)

3 0

1 year ago

Suppose the foreman had released the box from rest at a height of 0.25 m above the ground. What would the crate's speed be when

Arturiano [62]

Answer:

v = 2.21 m/s

Explanation:

The foreman had released the box from rest at a height of 0.25 m above the ground.

We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s$

So, its velocity at the bottom of the ramp is 2.21 m/s.

4 0

1 year ago

You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfo

dolphi86 [110]

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³ / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

= 2.33 x 10³ kg / m³ .

3 0

2 years ago

A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x fro

MArishka [77]

Answer:

Resistance = 3.35* $10^{-4}$ Ω

Explanation:

Since resistance R = ρ $\frac{L}{A}$

whereas $\rho(x) = a + bx^2$

resistivity is given for two ends. At the left end resistivity is $2.25* 10^{-8}$ whereas x at the left end will be 0 as distance is zero. Thus

$2.25*10^{-8} = a + b(0)^2\\ 2.25*10^{-8} = a + 0 \\2.25*10^{-8} = a$

At the right end x will be equal to the length of the rod, so $x = 1.50\\8.50*10^{-8} = (2.25*10^{-8}) + ( b* (1.50)^2 )\\8.50*10^{-8} - (2.25*10^{-8}) = b*2.25\\\frac{6.25*10^{-8}}{2.25} = b\\b = 2.77 *10^{-8}$

Thus resistance will be R = ρ $\frac{L}{A}$

where A = π $r^2$

so,

$R = \frac{8.50*10^{-8} * 1.50}{3.14*(1.10*10^{-2})^2} \\R=3.35 * 10 ^{-4}$

6 0

1 year ago