Describe the difference between red light and blue light.

To calculate the pressure at the entrance end of the pipe, we can apply Bernoulli's principle and the equation of continuity. First, we can find the speed of the liquid at the exit end of the pipe using the equation of continuity. Then, we can use Bernoulli's equation to calculate the pressure at the entrance end of the pipe. The pressure at the entrance end of the pipe is approximately 1.08 × 10^5 Pa.

To calculate the pressure at the entrance end of the pipe, we can apply Bernoulli's principle, which states that the total energy of a fluid flowing in a pipe is constant. Bernoulli's equation is given by:

P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2

In this equation, P1 and P2 are the pressures at the entrance and exit ends of the pipe, ρ is the density of the liquid, v1 and v2 are the speeds of the liquid at the entrance and exit ends, g is the acceleration due to gravity, and h1 and h2 are the heights of the liquid at the entrance and exit ends, respectively.

Using the given values, we can calculate:

First, we need to calculate v2 using the equation of continuity, which states that the mass flow rate is constant:

A1v1 = A2v2

Using the diameters of the pipe at the entrance and exit ends, we can find the areas:

Substituting the given values, we can solve for v2:

A1v1 = A2v2

[(π/4)(0.26^2)](1.06) = [(π/4)(0.05^2)]v2

v2 = [(π/4)(0.26^2)](1.06) / [(π/4)(0.05^2)]

v2 ≈ 22.22 m/s

Now that we have v2, we can substitute all the values into Bernoulli's equation and solve for P1:

P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2

Substituting the given values:

P1 + (1/2)(1136)(1.06^2) + (1136)(9.8)(0) = (1.2 × 101325) + (1/2)(1136)(22.22^2) + (1136)(9.8)(4.25)

Solving for P1:

P1 ≈ 1.08 × 10^5 Pa

Therefore, the pressure at the entrance end of the pipe is approximately 1.08 × 10^5 Pa.

Heat in a form of magnetic field energy can be calculated using: [tex]E_m = \frac{LI^2}{2}[/tex] where L is induction: [tex]L=\frac{\mu_0N^2A}{d}[/tex] and I electric current. The number of wraps N around coil is conditioned by the length of coil and thickness of wire. The more thick the wire the less wraps of wire around the coil while length will not change anything because it is the same. So if you make less wraps around the coil and construct the formula: [tex]E_m=\frac{\frac{\mu_0N^2A}{d}\cdot I^2}{2}[/tex] and simplify it to: [tex]E_m=\frac{\mu_0N^2AI^2}{2d}[/tex] which would mean that when your wire is thinner you produce more heat because you can make more wraps around the coil.

Hope this helps.

r3t40

Answer: the momentum is increased

Energy and momentum are conserved, this was proved by he American physicist Arthur H. Compton after his experiments related to the scattering of photons from electrons (Compton Effect), in which he also proved that photons do have momentum.

So, this momentum [tex]p[/tex] is given by the following expression:

[tex]p=\frac{h}{\lambda}[/tex]

Where [tex]h[/tex] is the Planck constant and [tex]\lambda[/tex] is the wavelength of the photon.

As we can see, the momentum is inversely proportional to to the wavelength. This means, if [tex]\lambda[/tex] decreases, [tex]p[/tex] increases.

I'm pretty that the answer would be D.

       Answer:

The location where the storm occurs

       Explanation:

The only difference between a cyclone and a hurricane is the location where the storm occurs. Hurricanes, cyclones and typhoons are the same thing, just different names are used for them depending on where they occur.

Mordancy.

Hurricanes are extremely violent cyclones, typically originating near the equator and moving north. Hurricanes are usually accompanied by heavy rains.

Cyclones are violent storms with winds that twist clockwise in the southern hemisphere, and counterclockwise in the northern hemisphere.  

Answer:

Explanation:

Before Thomson's discovery, atoms were believed according to the "Dalton's atomic theory" to be the smallest indivisible particle of any matter. This makes atoms the smallest unit of a matter.

Thomson in 1897, used the discharge tube to discover cathode rays which are today called electrons.

The discovery of electrons provided more light into the structure and nature of atoms. Atoms were now being seen in a different light as particles that are made up of other smaller sized particles.

Thomson through his experiment was able determine perfectly well the nature of the rays he saw emanating from the cathode. One of his findings shows that the rays are negatively charged and are repelled by negative charges.

The discovery of electrons further led to more works on the atom and other particles were discovered. Atoms were no longer seen as indivisible or the smallest particles of matter.

Answer:

40 cm behind the lens

Explanation:

Please refer to the attached picture for the ray tracing, where

f is the focal length

p is the location of the object

q is the location of the image

Each tick in the picture corresponds to 20 cm, so we can observe that

p = 40 cm

f = 20 cm

q = 40 cm

Also, q is inverted, real and same size as the object.

The location of the image, q, can also be verified by using the lens equation:

[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{20 cm}-\frac{1}{40 cm}=\frac{1}{40 cm}\\q = 40 cm[/tex]

The rocket's altitude [tex]y[/tex] at time [tex]t[/tex] is

[tex]y(t)=\left(800\dfrac{\rm km}{\rm h}\right)t[/tex]

so that after [tex]t=3\,\mathrm{min}[/tex], it will have traveled

[tex]y=\left(800\dfrac{\rm km}{\rm h}\right)(3\,\mathrm{min})=40\,\mathrm{km}[/tex]

The angle of elevation [tex]\theta[/tex] at time [tex]t[/tex] is such that

[tex]\tan\theta=\dfrac{y(t)}{13\,\rm km}[/tex]

At the moment when [tex]y(t)=30\,\mathrm{km}[/tex], this angle is

[tex]\tan\theta=\dfrac{30\,\rm km}{13\,\rm km}\implies\theta\approx66.57^\circ[/tex]

Differentiating both sides of the equation above gives

[tex]\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac1{13\,\rm km}\dfrac{\mathrm dy(t)}{\mathrm dt}[/tex]

and substituting the angle [tex]\theta[/tex] found above, and [tex]\dfrac{\mathrm dy(t)}{\mathrm dt}=800\dfrac{\rm km}{\rm h}[/tex], we get

[tex]\sec^2\theta66.57^\circ\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac1{13\,\mathrm km}\left(800\dfrac{\rm km}{\rm h}\right)[/tex]

[tex]\implies\boxed{\dfrac{\mathrm d\theta}{\mathrm dt}\approx9.73\dfrac{\rm rad}{\rm h}}[/tex]

Final answer:

Using related rates and the tangent function, the rate at which the angle between the telescope and the ground is increasing can be calculated 3 minutes after the rocket's launch.

Explanation:

To solve the problem of a rocket's angle of elevation changing as it travels vertically, we need to apply related rates in calculus. At the time given (3 minutes after lift-off), we should first convert the rocket's speed into meters per second (800 km/hr is approximately 222.22 m/s). Then, we can use the tangent function (tan(θ) = opposite/adjacent), where opposite is the height of the rocket and the adjacent is the distance from the observer to the launching pad (13 km). The rate of change of the angle can be found by differentiating the tangent function with respect to time and then using the chain rule to substitute for dh/dt (the rate at which the height of the rocket is changing).

After calculating the height of the rocket 3 minutes after launch (which would be 222.22 m/s * 180 s = 39999.6 m), we can find the rate of change of the angle as the rocket ascends by using the formula dθ/dt = (1/(1 + (h/13,000)2)) * (dh/dt)/(13,000), substituting h for 39999.6 and dh/dt for 222.22. The final calculation will give us the rate at which the angle between the telescope and the ground is increasing.

Answer:

Venus

Explanation:

Venus is the second plate in the solar system. It is a terrestrial planet and it is part of the inner rocky planets.

In Venus, it rains sulfuric acid but the rain never reaches the surface before it becomes evaporated. The acid forms from the combination of sulfur oxide and water in the atmosphere at a height of about 42km. As it condenses and falls, it becomes evaporated back at lower elevations. The surface is therefore protected from the sulfuric acid rain.

The sulfur oxide and water vapor must have been derived from volcanic activities in geologic times past.

Scientists cannot directly observe dark matter, but they know it exists by C, the way it exerts a gravitational pull on other matter. Most of the cosmic entities like galaxies do not have enough observable matter within them to logically exist, i.e. the amount of matter they have can’t hold the galaxy together.

1. [tex]8.91\cdot 10^{-7} m[/tex]

The wavelength of a wave is given by the formula

[tex]\lambda=\frac{v}{f}[/tex]

where

v is the speed of the wave

f is the frequency

For the electromagnetic wave in this problem,

[tex]f=2.30\cdot 10^{14}Hz[/tex] is the frequency

[tex]v=2.05\cdot 10^8 m/s[/tex] is the speed of the wave

Substituting into the equation, we find

[tex]\lambda=\frac{2.05\cdot 10^8 m/s}{2.30\cdot 10^{14}Hz}=8.91\cdot 10^{-7} m[/tex]

2.  [tex]22.1^{\circ}[/tex]

The angle of refraction can be found by using Snell's law:

[tex]n_i sin \theta_i = n_r sin \theta_r[/tex]

where

[tex]n_i = 1.00293[/tex] is the refractive index of the first medium (air)

[tex]n_r = 1.333[/tex] is the refractive index of the second medium (water)

[tex]\theta_i = 30.0^{\circ}[/tex] is the angle of incidence in air

Solving the equation for [tex]\theta_r[/tex], we find the angle of refraction of the light ray in water:

[tex]\theta_r = sin^{-1} (\frac{n_i sin \theta_i}{n_r})=sin^{-1} (\frac{(1.00293)(sin 30^{\circ})}{1.333})=22.1^{\circ}[/tex]

Answer:

14.4 L

Explanation:

Initially, the gas is at stp (standard conditions), which means

[tex]T_1 = 273 K\\p_1= 1.01 \cdot 10^5 Pa[/tex]

and its initial volume is

[tex]V_1 = 10 L = 0.010 m^2[/tex]

Later, the gas is heated to a final temperature of

[tex]T_2=512 C + 273 =785 K[/tex]

and the pressure is increased to

[tex]p_2 = 1520.0 mmHg \cdot \frac{1.01\cdot 10^5 Pa}{760 mmHg}=2.02\cdot 10^5 Pa[/tex]

So we can use the ideal gas equation to find the new volume, V2:

[tex]\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\\V_2 = \frac{p_1 V_1 T_2}{p_2 T_1}=\frac{(1.01\cdot 10^5 Pa)(0.010 L)(785 K)}{(2.02\cdot 10^5 Pa)(273 K)}=0.0144 m^2 = 14.4 L[/tex]

If 10.0 liters of oxygen at STP are heated to 512 °C and its pressure increased to 1520.0 mmHg, the new volume will be 14.4 L.

10.0 L (V₁) of oxygen is at standard temperature (T₁ = 273.15 K) and standard pressure (P₁ = 760.0 mmHg).

It is heated to 512 °C (T₂) and the pressure increased to 1520.0 mmHg (P₂). We will convert 512 °C to Kelvin using the following expression.

[tex]K = \° C + 273.15 = 512\° C + 273.15 = 785 K[/tex]

Then, we can calculate the new volume (V₂) using the combined gas law.

[tex]\frac{P_1 \times V_1 }{T_1} = \frac{P_2 \times V_2 }{T_2} \\\\V_2 = \frac{P_1 \times V_1 \times T_2 }{T_1 \times P_2} = \frac{760.0 mmHg \times 10.0 L \times 785 K }{273.15 K \times 1520.0 mmHg} = 14.4 L[/tex]

If 10.0 liters of oxygen at STP are heated to 512 °C and its pressure increased to 1520.0 mmHg, the new volume will be 14.4 L.

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Answer: Electromagnetic radiation

Explanation:

Electromagnetic radiation is a combination of oscillating electric and magnetic fields, which propagate through space carrying energy from one place to another.

To understand it better:

This radiation is spread thanks to the electromagnetic fields produced by moving electric charges and their sources can be natural or man-made.

It should be noted that the energy of electromagnetic radiation can vary and depending on its frequency it can be useful for various situations.

The statement best describes Kathy’s error is Nebulae formed around 109 y after hydrogen and helium formed. The correct option is third.

Universe is the whole cosmic system of matter and energy. The Earth, the Sun, the Moon, the planets, the stars along with the dust , gases, rocks forms the universe.

Kathy drew a timeline to show some of the major events that occurred during the evolution of the universe.

The universe consist of fully dark matter, interstellar gases or dust. The big bang is increasing just because in universe there is abundant of hydrogen and helium.

The Nebulae is formed around 109 years after hydrogen and helium were formed.

Thus, the correct option is third.

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Answer:

5.36 times bigger

Explanation:

The radius of the Earth is:

[tex]r_e = 6371 km[/tex]

so its diameter is

[tex]d_e = 2 r_2 = 12742 km[/tex]

The radius of Pluto is:

[tex]r_p = 1188 km[/tex]

So its diameter is

[tex]d_p = 2 r_p = 2376 km[/tex]

So the ratio between the two diameters is

[tex]\frac{d_e}{d_p}=\frac{12742 km}{2376 km}=5.36[/tex]

So the diameter of the Earth is 5.36 times bigger than the diameter of Pluto.

Earth's diameter is about 5.4 times larger than that of Pluto. Earth measures approximately 7,926 miles across, while Pluto has a diameter of about 1,473 miles.

The diameter of Earth is significantly larger than that of Pluto. Earth's equatorial diameter is approximately 7,926 miles (12,756 km), whereas Pluto's diameter is about 1,473 miles (2,370 km), making the Earth's diameter roughly 5.4 times larger than Pluto's.

This comparison illustrates the vast size differences between the planets in our solar system. While Jupiter's diameter is nearly 11 times that of Earth's, and Saturn's is about 9 times larger, Pluto remains one of the smallest bodies, now classified as a dwarf planet and far out into the solar system, with its distance from the Sun being over 30 astronomical blocks away.

(a) [tex]-1.48\cdot 10^{-5}N[/tex]

The electrostatic force exerted between the two sphere is given by:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where

k is the Coulomb's constant

q1, q2 are the charges on the two spheres

r is the separation between the centres of the two spheres

In this problem,

[tex]q_1 = 12\cdot 10^{-9} C\\q_2 = -23\cdot 10^{-9} C\\r = 0.41 m[/tex]

Substituting these values into the equation, we find the force

[tex]F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(12\cdot 10^{-9}C)(-23\cdot 10^{-9} C)}{(0.41 m)^2}=-1.48\cdot 10^{-5}N[/tex]

And the negative sign means the force is attractive, since the two spheres have charges of opposite sign.

(b) [tex]+1.62\cdot 10^{-6}N[/tex]

The total net charge over the two sphere is:

[tex]Q=q_1 +q_2 = 12\cdot 10^{-9}C+(-23\cdot 10^{-9}C)=-11\cdot 10^{-9} C[/tex]

When the two spheres are connected, the charge distribute equally over the two spheres (since they are identical, they have same capacitance), so each sphere will have a charge of

[tex]q=\frac{Q}{2}=\frac{-11\cdot 10^{-9}C}{2}=-5.5\cdot 10^{-9}C[/tex]

So the electrostatic force between the two spheres will now be

[tex]F=k\frac{q^2}{r^2}[/tex]

And substituting numbers, we find

[tex]F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(-5.5\cdot 10^{-9} C)^2}{(0.41 m)^2}=+1.62\cdot 10^{-6}N[/tex]

and the positive sign means the force is repulsive, since the two spheres have same sign charges.

The final charge on each sphere after contact will be −4 nC, which reflects the even distribution of the combined charges initially present on both spheres.

When two charged metal conducting spheres are brought into contact, their charges redistribute evenly between them because they are identical. This is because the total charge is conserved, and it will be shared equally when the two spheres touch.

If sphere A has a charge of −5 nC (nanoCoulombs) and sphere B has a charge of −3 nC, after they touch, they will both have the same charge. We calculate the final charge on each by adding the charges together and then dividing by two, which gives us (−5 nC + −3 nC)/2 = −4 nC on each sphere. The −4 nC corresponds to a specific number of electrons, where the charge of one electron is ≎1.6×10−19 C.

Answer:

3. +3.09 kJ

Explanation:

The change in internal energy of the gas is given by the 1st law of thermodynamics:

[tex]\Delta U=Q-W[/tex]

where

Q is the heat transferred to the gas

W is the work done by the gas

Here we have:

[tex]Q=+4.00 kJ[/tex] is the amount of heat transferred to the nitrogen

[tex]p=3.00 atm = 3.03\cdot 10^5 Pa[/tex] is the pressure of the gas

[tex]\Delta V=4.00 L-1.00 L=3.00 L = 0.003 m^3[/tex] is the change in volume of the gas

So the work done by the gas is

[tex]W=p\Delta V=(3.03\cdot 10^5 Pa)(0.003 m^3)=909 J = 0.91 kJ[/tex]

So, the change in internal energy of the gas is

[tex]\Delta U=4.00 kJ-0.91 kJ=+3.09 kJ[/tex]

20.november1998 im pretty sure

Answer:

D. 5.2Ω

Explanation:

Resistors in parallel have a net resistance of:

R = (1/R₁ + 1/R₂)⁻¹

The 2.0Ω resistor and the 3.0Ω resistor are in parallel, so the combination has a net resistance of:

R = (1/2.0 + 1/3.0)⁻¹

R = 1.2Ω

Resistors in series of a net resistance of:

R = R₁ + R₂

The 4.0Ω resistor is in series with the 2.0Ω/3.0Ω combination.  So the net resistance of all three is:

R = 1.2Ω + 4.0Ω

R = 5.2Ω

Answer:

conversion of mass into energy

Explanation:

Answer:

The distance between interference fringes increases.

Explanation:

In a double-slit diffraction pattern, the angular position of the nth-maximum in the diffraction patter (measured with respect to the central maximum) is given by

[tex]sin \theta = \frac{n \lambda}{d}[/tex]

where

[tex]\theta[/tex] is the angular position

[tex]\lambda[/tex] is the wavelength

d is the separation between the slits

In this problem, the separation between the slits decreases: this means that d in the formula decreases. As we see, the value of [tex]sin \theta[/tex] (and so, also [tex]\theta[/tex]) is inversely proportional to d: so, if the d decreases, then the angular separation between the fringes increases.

So, the correct answer is

The distance between interference fringes increases.

Answer:

-6.0 m/s, 10.4 m/s

Explanation:

To find the x- and y- components, we have to apply the formulas:

[tex]v_x = v cos \theta[/tex]

[tex]v_y = v sin \theta[/tex]

where

v = 12.0 m/s is the magnitude of the vector

[tex]\theta[/tex] is the angle between the direction of the vector and the positive x-axis

Here, the angle given is the angle above the negative x-axis; this means that the angle with respect to the positive x-axis is

[tex]\theta=180^{\circ} - 60^{\circ} = 120^{\circ}[/tex]

So, the two components are:

[tex]v_x = (12.0 m/s) cos 120^{\circ}=-6.0 m/s[/tex]

[tex]v_y = (12.0 m/s) sin 120^{\circ}=10.4 m/s[/tex]

The change of displacement with respect to time is defined as the velocity. The x component of the velocity will be -6.0 m/sec. while the velocity in the y-direction will be 10.4 m/sec.

The change of displacement with respect to time is defined as the velocity.

velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

[tex]V_x = V {cos\theta}[/tex]

[tex]V_y = V {sin\theta}[/tex]

[tex]\theta[/tex] = angle measured from the positive x-axis

In the above conditions, the angle is given from the - ve x-axis is 60 degrees. If it is measured from the + ve x-axis it will be 180- 60= 120 degrees.

Given velocity v = 12.0 m/sec

[tex]V_x = 12{cos20}[/tex] = -6.0 m/sec

[tex]V_y = 12 {sin120}[/tex] = 10.4 m/sec

Hence the x component of the velocity will be -6.0 m/sec. while the velocity in the y-direction will be 10.4 m/sec.

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A) See figure in attachment

There are 4 forces acting on the crate:

- The horizontal force F, pushing the crate to the right

- The frictional force [tex]F_f[/tex], acting in the opposite direction (to the left)

- The weight of the crate, [tex]W=mg[/tex], acting downward (with m being the mass of the crate and g the acceleration due to gravity)

- The normal reaction of the floor agains the crate, N, acting upward, and of same magnitude of the weight

The crate is in equilibrium (it is not moving), this means that the forces along the horizontal direction and along the vertical direction are balanced, so:

[tex]F=F_f\\N=W[/tex]

B) 490 N

In order to make the crate start sliding along the floow, the horizontal push must overcome the maximum force of static friction, which is given by

[tex]F_f = \mu_s N[/tex]

where

[tex]\mu_s=0.56[/tex] is the coefficient of static friction

N is the normal reaction

The normal reaction is equal to the weight of the crate, so

[tex]N=W=875 N[/tex]

and so, the maximum force of static friction is

[tex]F_f = (0.56)(875 N)=490 N[/tex]

The magnitude of the minimum force you need to exert on the crate to make it start sliding along the floor is 490 N.

The given parameters;

The normal force on the crate is calculated as follows;

Fₙ = W = 875 N

The static frictional force on the crate at rest;

[tex]F_s = \mu_s F_n\\\\F_s = 0.56 \times 875 \\\\F_s = 490 \ N[/tex]

Before the crate will move this static frictional force must be overcome.

Thus, the magnitude of the minimum force you need to exert on the crate to make it start sliding along the floor is 490 N.

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(a) [tex]\frac{4}{3}v[/tex]

Let's write the law of conservation of momentum for the collision:

[tex]m_A u_A + m_B u_B = m_A v_A + m_B v_B[/tex] (1)

where u refers to the initial velocities and v refers to the velocity after the collision.

The problem gives us the following information:

- car B is one-half as massive as car A, so we can write:

[tex]m_A = 2m\\m_B = m[/tex]

- car B is initiall stationary:

[tex]u_B=0[/tex]

- After the collision, car A moves in the same direction as before with a speed v/3:

[tex]u_A = v\\v_A = \frac{1}{3}u_A=\frac{1}{3}v[/tex]

So we can rewrite (1) as

[tex](2m) v = (2m) \frac{v}{3}+mv_B[/tex]

and solving for [tex]v_B[/tex], we find the final speed of car B:

[tex]v_B = \frac{6v-2v}{3}=\frac{4}{3}v[/tex]

(b) Elastic

To find if the collision is elastic or inelastic, we have to check if the total kinetic energy has been conserved or not.

The total kinetic energy before the collision is:

[tex]K_i = \frac{1}{2}m_A u_A^2 = \frac{1}{2}(2m)(v)^2=mv^2[/tex]

The total kinetic energy after the collision is:

[tex]K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2=\frac{1}{2}(2m)(\frac{v}{3})^2+\frac{1}{2}m(\frac{4}{3}v)^2=\frac{1}{9}mv^2+\frac{8}{9}mv^2=mv^2[/tex]

The total kinetic energy has been conserved: so, the collision is elastic.

Applying the law of conservation of momentum to the collision leads to finding the second car's final speed as 4v/3. This is an inelastic collision, as kinetic energy is not conserved.

The scenario presented involves the concept of momentum conservation. The total momentum before the collision equals the total momentum after the collision. Before the collision, the stationary car, having a mass equal to half of the moving car and no velocity, has zero momentum. The moving car has initial momentum equal to its mass (m) times its velocity (v).

After the collision, the first car's final momentum is m*(v/3), and the second car's final momentum is its mass (m/2) times its unknown final velocity (v2). By setting the total initial momentum equal to the total final momentum, we find v2 = 4v/3.

To answer the second part, this is an inelastic collision, because in this type of collision, energy is not conserved even though momentum is conserved. The first car loses speed, meaning it loses kinetic energy, which is not fully gained by the second car.

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The dwarf planet Pluton, has the most eccentric orbit (more elliptical and elongated) of all the planets, and as a consequence its orbit is "intersected" by the orbit of Neptune.

However, despite this intersection, there is no collision risk between these two bodies, since the orbit of Pluto is located in an orbital plane different from that of the other planets and therefore different from that of Neptune, its nearest neighbor. In addition, the orbit of Pluton is inclined [tex]17\°[/tex] on the the ecliptic plane (plane where the other planets move around the Sun).

1) [tex]1.11\cdot 10^{-7} J[/tex]

The capacitance of a parallel-plate capacitor is given by:

[tex]C=\frac{\epsilon_0 A}{d}[/tex]

where

[tex]\epsilon_0[/tex] is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

[tex]r=\frac{2.0 cm}{2}=1.0 cm=0.01 m[/tex]

so the area is

[tex]A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2[/tex]

While the separation between the plates is

[tex]d=0.50 mm=5\cdot 10^{-4} m[/tex]

So the capacitance is

[tex]C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F[/tex]

And now we can find the energy stored,which is given by:

[tex]U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J[/tex]

2) 0.71 J/m^3

The magnitude of the electric field is given by

[tex]E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m[/tex]

and the energy density of the electric field is given by

[tex]u=\frac{1}{2}\epsilon_0 E^2[/tex]

and using

[tex]E=4\cdot 10^5 V/m[/tex], we find

[tex]u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3[/tex]

Answer:

0.75 m

Explanation:

Let's call the distance between the bulb and the mirror x.

The bulb and the length of the mirror form a triangle.  The mirror and the illuminated area on the floor form a trapezoid.  If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle.  This triangle and the small triangle are similar.  So we can say:

x / 0.4 = (3 + x) / 2

Solving for x:

2x = 0.4 (3 + x)

2x = 1.2 + 0.4 x

1.6 x = 1.2

x = 0.75

So the bulb should located no more than 0.75 m from the mirror.

Answer:

They are found in wet and dry areas

Explanation:

not all house fires can be prevented.

GFCI outlets are found in wet areas and prevent electrocution if you touch a wet appliance. They work by comparing the currents in the wires and interrupt the circuit if a leakage current is detected. They do not specifically prevent house fires and can be found in dry areas, but are especially needed in wet areas.

The true statements about a GFCI (Ground Fault Circuit Interrupter) outlet are: B. GFCI outlets are found in wet areas such as kitchens and bathrooms, and C. GFCI outlets prevent electrocution if you are touching a wet appliance. A GFCI outlet compares the current in the live/hot and the neutral wires, triggering an interrupt if any leakage current is detected, such as when a person is accidentally creating a path for the current by touching a wet appliance.

This prevents electrical shock. Statement A is not entirely accurate because GFCI outlets are more about preventing electric shock than house fires. Statement D is misleading as GFCI outlets can be found anywhere but are specifically important in wet areas where the risk of accidental grounding through a person is higher.

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I think the answer is d

Answer:

Approximately 1.5 × 10⁵ Pa.

Explanation:

In simple words, pressure [tex]P[/tex] is the normal force per unit area:  

[tex]\displaystyle P = \frac{F(\text{Normal Force})}{A}[/tex],

where

What will be the size of the normal force?

Weight [tex]W[/tex] of the elephant, which is the same as the gravitational pull on the elephant, will be:

[tex]W = m \cdot g = 6000\times 9.81 = 58860\;\text{N}[/tex],

where

The elephant stands on solid ground. It's not accelerating upward or downward. Forces on the elephant are balanced. As a result, the size of the normal force on the elephant will be the same as that of the earth's gravitational pull on the elephant (not the case in an accelerating elevator.)

[tex]F(\text{Normal Force}) = W = 58860\;\text{N}[/tex].

Each elephant got four feet. The area of each foot is [tex]0.10\;\text{m}^{2}[/tex]. The total contact area will be [tex]4 \times 0.10 = 0.40\;\text{m}^{2}[/tex].

Apply the formula. Make sure both the normal force and the contact area are in SI units:

If that's the case, the pressure result will be in the unit newtons per square meters [tex]\text{N}\cdot \text{m}^{-2}[/tex] or equivalently, Pascals [tex]\text{Pa}[/tex].

[tex]\displaystyle P = \frac{F(\text{Normal Force})}{A} = \frac{58860\;\text{N}}{0.40\;\text{m}^{2}} \approx 1.5\times 10^{5}\;\text{N}\cdot\text{m}^{-2} = 1.5\times 10^{5}\;\text{Pa}[/tex].

The weight of the elephant is the force exerted on the ground, which can be calculated using the equation Force = mass x acceleration due to gravity. The pressure exerted by the feet of the elephant on the ground is determined by dividing the force by the area of each foot.

When determining the pressure exerted by the feet of an elephant on the ground, we need to calculate the force using the weight of the elephant. The weight of an object is the force exerted due to gravity. We can use the equation:

Force = mass x acceleration due to gravity

In this case, the mass of the elephant is 6000 kg and the acceleration due to gravity is approximately 9.8 m/s^2. So the force exerted by the elephant is:

Force = 6000 kg x 9.8 m/s^2 = 58,800 N

To find the pressure, we divide the force by the area of each foot. In this case, each foot has an area of 0.10 square meter. So the pressure exerted by the feet of the elephant on the ground is:

Pressure = Force / Area = 58,800 N / 0.10 m^2 = 588,000 Pa

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