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ludmilkaskok [199]

3 years ago

5

An ink-jet printer steers charged ink drops vertically. Each drop of ink has a mass of 10-11 kg, and a charge due to 500,000 ext

ra electrons. It goes through two electrodes that gives a vertical acceleration of 104 m/s2. The deflecting electric field is _____ MV/m.

1 answer:

Sphinxa [80]3 years ago

4 0

Answer:

E = 1.25 MV / m

Explanation:

For this exercise let's use Newton's second law

F = m a

where the force is electric

F = q E

we substitute

q E = m a

E = m a / q

indicate there are 500,000 excess electrons

q = 500000 e

q = 500000 1.6 10⁻¹⁹

q = 8 10⁻¹⁴ C

the mass is m = 10⁻¹¹ kg and the acceleration a = 10⁴ m / s²

let's calculate

E = 10⁻¹¹ 10⁴ / 8 10⁻¹⁴

E = 0.125 10⁷ V / m = 1.25 10⁶ V / m

E = 1.25 MV / m

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

An eagle is flying horizontally at a speed of 3.80 m/s when the fish in her talons wiggles loose and falls into the lake 3.90 m

Dovator [93]

Answer:

the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal

Explanation:

initial veetical speed V₀y=0

Horizontal speed Vx = Vx₀= 3.80m/s

Vertical drop height= 3.90m

Let Vy = vertical speed when it got to the water downward.

g= 9.81m/s² = acceleration due to gravity

From kinematics equation of motion for vertical drop

Vy²= V₀y² +2 gh

Vy²= 0 + ( 2× 9.8 × 3.90)

Vy= √76.518

Vy=8.747457

Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below

V= √Vy² + Vx²

V=√3.80² + 8.747457²

V=9.537m/s

The angle can also be calculated as

θ=tan⁻¹(Vy/Vx)

tan⁻¹( 8.747457/3.80)

=66.52⁰

the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal

6 0

3 years ago

Science whoever gets this will get a brainlest

Zanzabum

Answer:

I got it right xd

Explanation:

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6 0

3 years ago

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When you push a 2.00 kg book resting on a tabletop it takes 4.60 N to start the book sliding. What is the coefficient of static

natali 33 [55]

The coefficient of static friction is 0.234.

Answer:

Explanation:

Frictional force is equal to the product of coefficient of friction and normal force acting on any object.

So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.

Normal force = mass * acceleration due to gravity

Normal force = 2 * 9.8 = 19.6 N.

And the frictional force is given as 4.6 N, then

$Coefficient of static friction = Frictional force/Normal force$

Coefficient of static friction = 4.6 N / 19.6 N = 0.234

So the coefficient of static friction is 0.234.

3 0

3 years ago

A cyclist moves at a constant speed of 5 m/s if the cyclist does not accelerate during the next 20 seconds he will travel at?

Verizon [17]

5 m/s because the speed is constant

8 0

3 years ago