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s2008m [1.1K]
3 years ago
6

In which point the electric intensity of a sphere is maximum​

Physics
1 answer:
g100num [7]3 years ago
7 0
Here, as the charge is uniformly distributed in the sphere, we will consider s as an area of the sphere which is, s=4πr2 and r is radius of the gaussian surface shown in the figure above. From this, it can be seen that
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Mr. Mann explained that when light passes from air into the water, the light waves bend, or ___________, and the pencil appears
posledela

Answer:

the light waves bend or REFRACT

Explanation:

4 0
3 years ago
What does an alpha particle consist of? one proton and two neutrons two protons and two neutrons two protons and one electron on
kolbaska11 [484]

They are helium nuclei, which consist of two protons and two neutrons. The net spin on an alpha particle is zero. They result from large, perilous atoms via a process called alpha decay.

<h3>What is helium nuclei?</h3>
  • The nucleus of the helium atom also understood as the α-particle, includes two protons and two neutrons, encompassed by two electrons.
  • Alpha particles are helium nuclei with two protons and two neutrons attached. The development of their high mass and an electrical charge is their inability to infiltrate as deep as other particles such as protons and electrons.
  • Particle beams contain α (alpha)-particles, β (beta)-particles, neutron beams, etc. α-particles are helium middles consisting of two protons and two neutrons that have lived removed at high speed, while β-particles are electrons removed from a nucleus. Particle shafts also include neutron beams and proton beams.

To learn more about helium nuclei, refer to:

brainly.com/question/26226232

#SPJ4

3 0
2 years ago
Find 4-5 animals and classify them into Prey, Predator, Carnivore, Omnivore, Herbivore.​
Sveta_85 [38]

Answer:

Koala, leopard, lion, cat, dog. Koala is an herbivore, leopard is a predator, lion is a predator, cat is a obligate carnivore, dogs are omnivores.

Explanation:

7 0
2 years ago
A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a
Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

Z_{R} = R

R = resistance

Z_{L} = jωL

ω = voltage source angular frequency, L = inductance

Z_{C} = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

Z_{R} = 217Ω

Z_{L} = j(220)(0.875) = j192.5Ω

Z_{C} = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = Z_{R} + Z_{L} + Z_{C}

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

v_{R} = I×Z_{R}

v_{L} = I×Z_{L}

v_{C} = I×Z_{C}

Given values:

I = (0.0569∠1.147+220t)A

Z_{R} = 217Ω = (217∠0)Ω

Z_{L} = j192.5Ω = (192.5∠π/2)Ω

Z_{C} = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

v_{R} = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

v_{L} = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

v_{C} = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

v_{R} = (12.35∠1.147+220t)V

v_{L} = (10.95∠2.718+220t)V

v_{C} = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

v_{R} = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

v_{L} = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

v_{C} = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0
3 years ago
What does a mechanical wave always travel through?
jarptica [38.1K]
A mechanical wave<span> requires an initial energy input. Once this initial energy is added, the </span>wave travels through<span> the medium until all its energy is transferred.</span>
6 0
3 years ago
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