Simon draws a diagram to illustrate the law of reflection.



Which best explains how Simon can correct the error in his diagram?

Draw the incident ray so it is longer than the reflected ray.
Make the angle of reflection and the angle of incidence equal.
Switch the labels for incident ray and reflected ray.
Make the angle of incidence larger than the angle of reflection.

Answer:

Explanation:

The weight of the 4kg mass equals the force of the spring.

mg = kx

(4 kg) (9.8 m/s²) = k (0.025 m)

k = 1568 N/m

If an additional 1.5 kg is added, the spring stretches to:

(4 kg + 1.5 kg) (9.8 m/s²) = (1568 N/m) x

x = 0.034 m

x = 3.4 cm

If the 4 kg is removed, the spring stretches to:

(1.5 kg) (9.8 m/s²) = (1568 N/m) x

x = 0.0094 m

x = 0.94 cm

The work done to stretch a spring a distance x is:

E = 1/2 k x²

E = 1/2 (1568 N/m) (0.04 m)²

E = 1.3 J

Answer:

1/2

Explanation:

The energy stored in a capacitor is given by:

[tex]U=\frac{1}{2}CV^2[/tex]

where

C is the capacitance

V is the potential difference

For capacitor 1, we have

[tex]U_1=\frac{1}{2}C_1V_1^2[/tex]

Capacitor 2 has

[tex]C_2 = \frac{C_1}{2}[/tex] (half the capacitance of capacitor 1)

[tex]V_2 = 2 V_1[/tex] (twice the potential difference of capacitor 1)

So the energy of capacitor 2 is

[tex]U_2=\frac{1}{2}C_2V_2^2=\frac{1}{2}(\frac{C_1}{2})(2V_1)^2=C_1 V_1^2[/tex]

So, the ratio between the two energies is

[tex]\frac{U_1}{U_2}=\frac{\frac{1}{2}C_1 V_1^2}{C_1 V_1^2}=\frac{1}{2}[/tex]

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.

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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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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(a)

For this part of the problem, we can ignore the horizontal motion of the engine and consider only the vertical motion.

The vertical position of the engine at time t is given by

[tex]y(t) = h - \frac{1}{2}gt^2[/tex]

where

h is the initial altitude of the airplane

g = 9.81 m/s^2 is the acceleration due to gravity

t is the time

Since the engine takes 30 seconds to hit the ground, t = 30 s when y(t) = 0 (the ground). Substituting into the equation, we find h:

[tex]0 = h - \frac{1}{2}gt^2\\h= \frac{1}{2}gt^2 = \frac{1}{2}(9.81 m/s^2)(30 s)^2=4,415 m \sim 4.5 km[/tex]

(b)

For this part of the problem, we can ignore the vertical motion and consider the horizontal motion only.

Since the engine travels at constant speed along the horizontal direction:

v = 280 m/s

its horizontal position after time t is given by

[tex]x(t) = v t[/tex]

If we substitute

t = 30 s

which is the total duration of the fall, we can find the horizontal distance covered by the airplane during this time:

[tex]x(t) = (280 m/s)(30 s)=8,400 m[/tex]

(c) The engine will be exactly 4.5 km under the plane

Here we have:

- the airplane is moving horizontally, at 4.5 km of altitude, at constant velocity of 280 m/s

- The engine moves both horizontally, also with a horizontal velocity of 280 m/s, and vertically, with acceleration 9.81 m/s^2 towards the ground

Both the plane and the engine moves with same horizontal velocity, so they cover the same horizontal distance (8400 m) during the 30 seconds. The only difference is that the engine falls down approx. 4.5 km, so the engine will be 4.5 km under the plane, when it hits the ground.

The airplane is 4.5 km high and the horizontal distance that the aircraft engine moves during its fall is 8400 m.

To find the height if the plane from ground, we use:

H = ut + (1/2)gt²

Where u = initial velocity = 0 m/s, t = time = 30 s, g = acceleration due to gravity = 10 m/s²

H = 0(30) + (1/2)(10)(30)²

H = 4500 m = 4.5 km

The horizontal distance (R) is given by:

R = ut = 280(30) = 8400 m

Let d represent the distance from the engine to the air plane at the ground, hence:

d² = 8400² + 4500²

d = 9529 m

The airplane is 4.5 km high and the horizontal distance that the aircraft engine moves during its fall is 8400 m.

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The decibel rating of sound source B is 80 dB. So, the decibel rating of sound source B would be 50 dB + 10 * log10(1000) = 50 dB + 10 * 3 = 50 dB + 30 dB = 80 dB

The decibel rating of sound source B can be calculated by taking the decibel rating of sound source A and adding 10 times the logarithm base 10 of the intensity ratio between sound source B and A.

Since sound source B is 1000 times more intense than sound source A, the intensity ratio is 1000.

So, the decibel rating of sound source B would be 50 dB + 10 * log10(1000) = 50 dB + 10 * 3 = 50 dB + 30 dB = 80 dB.

Electromagnetic Spectrum

The electromagnetic spectrum is the infinite range of frequencies of electromagnetic radiation, but what does frequency have to do with electromagnetic radiation? Well, electromagnetic radiation travels in waves and frequency [tex]f[/tex] tells us how many cycles per second it has a particular wave. Another important term is wavelength [tex]\lambda[/tex]  that measures the distance from one crest to the next in a wave. Although you can find waves having different frequency and wavelength, in vacuum, however, there is a relationship for all of them:

[tex]c=\lambda f[/tex]

Where:

[tex]c[/tex] is the speed of light

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

[tex]f[/tex] is the frequency

But what are those waves? Radio, Microwave, Infrared, Ultraviolet, X rays, Gamma rays and even visible light are part of the electromagnetic spectrum.

Answer:

16 times

Explanation:

The international space station orbits the earth every 90 minutes. That means the international space station will orbit the earth 16 times within a 24 hour period.

Final answer:

The International Space Station orbits Earth once every 90 minutes in a Low Earth Orbit, circling the planet around 16 times per day, whereas geostationary satellites orbit once every 24 hours at a fixed point above the Earth's surface.

Explanation:

The International Space Station (ISS) orbits Earth approximately once every 90 minutes. It is in a type of orbit known as Low Earth Orbit (LEO), which ranges in altitude from 160 km to 2,000 km above Earth's surface, with the ISS specifically orbiting at around 370 km in the thermosphere. This rapid orbit allows the ISS to circumnavigate Earth around 16 times per day. In contrast, satellites in a geostationary orbit, such as weather and communications satellites, orbit at a much higher altitude of 36,000 km and take exactly 24 hours to complete one orbit. This synchronized period with Earth's rotation keeps them stationary over one specific location on the planet's surface.

Answer:

Zero Kelvin

Explanation:

The average kinetic energy of the particles in a gas is related to the absolute temperature of the gas by (for an ideal monoatomic gas):

[tex]E_K = \frac{3}{2}kT[/tex]

where

k is the Boltzmann constant

T is the absolute temperature

The average kinetic energy is the energy possessed by the particles due to their motion; we see that this energy becomes zero when T = 0, which means when the substance reaches a temperature of zero Kelvin. Therefore, this means that at this temperature all the particles stop moving.

Absolute zero is the temperature at which all molecular motion ceases, defined as 0 Kelvin. It is a theoretical limit that has never been fully reached but approximated in laboratory conditions. The temperature of a substance at this point is directly proportional to the average kinetic energy of its particles, which would be zero at absolute zero.

The temperature at which all molecular motion stops is known as absolute zero. Absolute zero is theoretically the lowest possible temperature where the internal energy of a system is minimal due to the cessation of molecular motion. This temperature is defined as 0 Kelvin (K), which is equivalent to -273.15°C or -459.67°F. In the Kelvin scale, which is based on molecular motion, the temperature of a substance is directly proportional to the average kinetic energy of its particles. While absolute zero has never been achieved in practice, scientists have managed to reach temperatures extremely close to this theoretical limit in laboratory settings.

It is important to correct the inaccurate statement (c) that the speed of particles increases to a maximum at absolute zero. In fact, it is point (d) which correctly states that the internal energy approaches zero because the speed of particles decreases to zero at this temperature.

Answer:

A = g (λ / 2π)² μ / Ts

Explanation:

The ant becomes "weightless" when its acceleration is equal to gravity.  The motion of the ant is a sinusoidal wave:

y = A sin(ωt)

By taking the derivative twice, we can find the acceleration of the ant:

y' = Aω cos(ωt)

y" = -Aω² sin(ωt)

The maximum acceleration occurs when sine is 1.  We want this to happen at a = -g.

-g = -Aω² (1)

A = g / ω²

Angular frequency is 2π times the normal frequency:

ω = 2πf

A = g / (2πf)²

Frequency is velocity divided by wavelength:

f = v / λ

A = g / (2πv / λ)²

A = g (λ / 2π)² / v²

Velocity of a wave in a string with tension Ts and linear density μ is:

v = √(Ts / μ)

Therefore:

v² = Ts / μ

Plugging in:

A = g (λ / 2π)² / (Ts / μ)

A = g (λ / 2π)² μ / Ts

The minimum wave amplitude in terms of Ts, μ, λ, and g is;

A = λ²gμ/(4π²Ts)

        Since we are told that the large ant stands on a rope that is stretched, then the position motion of the ant would be represented by a sinusoidal wave:

y = A sin(ωt)

         We are told that the ant becomes weightless when its acceleration is equal to gravity. Let us find the acceleration which is the second derivative of the position equation to get;

y' = Aω cos(ωt)

y" = -Aω² sin(ωt)

Acceleration; a = -Aω² sin(ωt)

          Now, the maximum acceleration will occur when sin(ωt) = 1 .  This will happen when a = -g.  

Thus;

-g = -Aω²

A = g/ω²

         Formula for angular frequency in terms of velocity and wavelength is;

ω = 2πv/ λ

Thus;

A = g/(2πv/λ)²

A = λ²g/(2λπ)²v²

        Now, the formula for velocity of the wave in terms of tension Ts and linear density μ is:

v = √(Ts/μ)

Thus;

v² = Ts/μ

Thus;

A = λ²g/((2π)²(Ts/μ))

A = λ²gμ/(4π²Ts)

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Answer:

E) 1500 N

Explanation:

There are two ways to solve this: energy equations or kinematics.

First I'll use energy equations.

All of Santa's energy is converted to work by friction.

Initial energy = final energy + work

PE = W

mgh = Fd

(120 kg) (9.8 m/s²) (9 m + 2 m) = F (9 m)

F = 1437 N

Using kinematics, the velocity Santa reaches when he reaches the chimney is:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2 (-9.8 m/s²) (9 m - 11 m)

v = -6.26 m/s

Then he starts decelerating down the chimney.  Finding the acceleration:

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (-6.26 m/s)² + 2 a (0 m - 9 m)

a = 2.18 m/s²

Sum of the forces acting on Santa:

∑F = ma

F - W = ma

F = W + ma

F = mg + ma

F = m (g + a)

F = (120 kg) (9.8 m/s² + 2.18 m/s²)

F = 1437 N

Rounded to 2 sig-figs, that's 1400 N, which isn't one of the choices.  But if we use 10 m/s² for g instead of 9.8 m/s², we get F = 1467 N, which rounds to 1500 N.

E) 1500 N

Refraction consists in the bending of light or the change in its direction when passing through a medium with a refractive index [tex]n[/tex] different from the other medium.  

Being its equation as follows:

[tex]n=\frac{c}{v}[/tex]

Where  [tex]c[/tex] is  the speed of light in vacuum and [tex]v[/tex] its speed in the other medium.

Now, writing this equation with the known values, we will have Z's index of refraction:

[tex]n=\frac{3(10)^{8} m/s}{2.5(10)^{8} m/s}[/tex]

[tex]n=1.2[/tex]

The index of refraction of material Z is calculated to be approximately 1.2 using the given speeds of light in a vacuum and in material Z.

To determine the index of refraction of material Z, we can use the formula for the index of refraction, n:

[tex]n = c / v[/tex]

Where:

We substitute the given values into the formula:

[tex]n = 3.00\times 10^8 m/s / 2.5\times 10^8 m/s[/tex]

By performing the division:

[tex]n = 1.2[/tex]

Therefore, the index of refraction of material Z is approximately 1.2.

A) [tex]C_3 < C_1 < C_2[/tex]

The capacitance of a parallel-plate capacitor is given by

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

where

A is the plate area

d is the plate separation

Here we have:

- Capacitor 1: plate area A, plate separation d

capacitance: [tex]C_1=\epsilon_0 \frac{A}{d}[/tex]

- Capacitor 2: plate area 2A, plate separation d

capacitance: [tex]C_2=\epsilon_0 \frac{2A}{d} = 2C_1[/tex]

- Capacitor 3: plate area A, plate separation 2d

capacitance: [tex]C_3=\epsilon_0 \frac{A}{2d}=\frac{C_1}{2}[/tex]

So ranking the three capacitor from least to greatest capacitance we have:

[tex]C_3 < C_1 < C_2[/tex]

2. [tex]V_2 < V_1 < V_3[/tex]

The three capacitors have same amont of charge, Q.

The potential difference between the plates on each capacitor is given by

[tex] V = \frac{Q}{C}[/tex]

so here we have

- Capacitor 1: [tex]C = C_1[/tex]

Potential difference: [tex] V_1 = \frac{Q}{C_1}[/tex]

- Capacitor 2: [tex]C = 2C_1[/tex]

Potential difference: [tex] V_2= \frac{Q}{2C_1}=\frac{ V_1}{2}[/tex]

- Capacitor 3: [tex]C = \frac{C_1}{2}[/tex]

Potential difference: [tex] V_3 = \frac{Q}{C_1/2}=2 V_1 [/tex]

So ranking the three capacitor from least to greatest potential difference we have:

[tex]V_2 < V_1 < V_3[/tex]

C. [tex]E_2 < E_1 = E_3[/tex]

The electric field magnitude between the plates of a capacitor is given by

[tex]E=\frac{V}{d}[/tex]

where V is the potential difference between the plates and d is the plate separation

So here we have

- Capacitor 1: potential difference [tex]V_1[/tex], plate separation d

electric field: [tex]E_1 = \frac{V_1}{d}[/tex]

- Capacitor 2: potential difference [tex]\frac{V_1}{2}[/tex], plate separation d

electric field: [tex]E_2=\frac{V_1/2}{d} =\frac{V_1}{2d}= \frac{E_1}{2}[/tex]

- Capacitor 3: potential difference [tex]2V_1[/tex], plate separation 2d

electric field: [tex]E_3=\frac{2 V_1}{2d} =\frac{V_1}{d}= E_1[/tex]

So ranking the three capacitor from least to greatest electric field we have:

[tex]E_2 < E_1 = E_3[/tex]

D. [tex]U_2 < U_1 < U_3[/tex]

The energy stored in a capacitor is

[tex]U=\frac{1}{2}QV[/tex]

where Q is the same for the three capacitors

Here we have

- Capacitor 1: potential difference [tex]V_1[/tex]

energy: [tex]U_1 = \frac{1}{2}QV_1[/tex]

- Capacitor 2: potential difference [tex]\frac{V_1}{2}[/tex]

energy: [tex]U_2 = \frac{1}{2}Q\frac{V_1}{2}=\frac{U_1}{2}[/tex]

- Capacitor 3: potential difference [tex]2V_1[/tex]

energy: [tex]U_3 = \frac{1}{2}Q(2 V_1)=2 U_1[/tex]

So ranking the three capacitor from least to greatest energy we have:

[tex]U_2 < U_1 < U_3[/tex]

E. [tex]u_2 < u_1 = u_3[/tex]

The energy density in a capacitor is given by

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

where E is the electric field strength

Here we have

- Capacitor 1: electric field [tex]E_1[/tex]

Energy density: [tex]u_1=\frac{1}{2}\epsilon_0 E_1^2[/tex]

- Capacitor 2: electric field [tex]\frac{E_1}{2}[/tex]

energy density: [tex]u_2=\frac{1}{2}\epsilon_0 (\frac{E_1}{2})^2=\frac{E_1}{4}[/tex]

- Capacitor 3: electric field [tex]E_1[/tex]

Energy density: [tex]u_3=\frac{1}{2}\epsilon_0 E_1^2[/tex]

So ranking the three capacitor from least to greatest energy density we have:

[tex]u_2 < u_1 = u_3[/tex]

Capacitor 2 has the highest capacitance, potential difference, energy stored, and energy density, followed by Capacitor 1, and then Capacitor 3.

Part A: The formula for capacitance is C = εA/d, where C is the capacitance, ε is the permittivity of the material between the plates, A is the plate area, and d is the plate separation. Comparing the three capacitors, Capacitor 1 has a capacitance of C1 = εA/d, Capacitor 2 has a capacitance of C2 = ε(2A)/d, and Capacitor 3 has a capacitance of C3 = εA/(2d). Since C2 = 2C1 and C3 = 1/2C1, Capacitor 2 has the highest capacitance, followed by Capacitor 1, and then Capacitor 3.

Part B: The potential difference between the plates of a capacitor is given by V = Q/C, where V is the potential difference, Q is the charge stored on the plates, and C is the capacitance. Since all three capacitors store the same amount of charge, their potential differences are directly proportional to their capacitances. Therefore, Capacitor 2 has the highest potential difference, followed by Capacitor 1, and then Capacitor 3.

Part C: The electric field magnitude between the plates of a capacitor is given by E = V/d, where E is the electric field magnitude and d is the plate separation. Comparing the three capacitors, Capacitor 1 has an electric field magnitude of E1 = V/d, Capacitor 2 has an electric field magnitude of E2 = (2V)/d, and Capacitor 3 has an electric field magnitude of E3 = V/(2d). Since E2 = 2E1 and E3 = 1/2E1, Capacitor 2 has the highest electric field magnitude, followed by Capacitor 1, and then Capacitor 3.

Part D: The energy stored in a capacitor is given by U = (1/2)CV^2, where U is the energy stored, C is the capacitance, and V is the potential difference. Comparing the three capacitors, Capacitor 1 has an energy stored of U1 = (1/2)C1V^2, Capacitor 2 has an energy stored of U2 = (1/2)C2V^2, and Capacitor 3 has an energy stored of U3 = (1/2)C3V^2. Since C2 > C1 > C3, Capacitor 2 has the highest energy stored, followed by Capacitor 1, and then Capacitor 3.

Part E: The energy density of a capacitor is given by u = U/Vd, where u is the energy density, U is the energy stored, V is the volume between the plates, and d is the plate separation. Comparing the three capacitors, Capacitor 1 has an energy density of u1 = U1/(Vd), Capacitor 2 has an energy density of u2 = U2/(2Vd), and Capacitor 3 has an energy density of u3 = U3/(2Vd). Since u2 = 2u1 and u3 = 1/2u1, Capacitor 2 has the highest energy density, followed by Capacitor 1, and then Capacitor 3.

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Jupiter's moon Ganymede is the largest satellite in the solar system. Larger than Mercury and Pluto, and only slightly smaller than Mars, it would easily be classified as a planet if were orbiting the sun rather than Jupiter

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Lever arm is defined as the "perpendicular distance that exists from the rotation axis to the line of action of the force". Being its magnitude [tex](Nm)[/tex] related to the application of the force that produces a torque.

To understand it better:

The lever arm is an effective distance to apply a force with respect to a certain point, and this distance serves as an effective applied force amplification factor.

So, the greater the lever arm, the greater the torque and the lesser the force we have to apply.

The perpendicular distance between the applied force and the center of rotation is known as the Lever Arm or Torque Arm and plays a significant role in calculating Torque, which is the measure of the force's effectiveness in switching or accelerating rotation.

The perpendicular distance between the applied force and the center of rotation is referred to as the Lever Arm or Torque Arm. This lever arm is crucial in calculating the Torque, which is defined as the product of the magnitute of the force and the perpendicular distance (Lever Arm) from the axis to the line upon which the force vector lies. Mathematically, it is represented as ||| = =r₁ F.

Therefore, the lever arm designs the effectiveness of a force to instigate a change or acceleration in rotation. An interesting connotation is that the greater the length of the lever arm, the more dramatic the angular acceleration will be.

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Friction, engine thrust, normal reaction and weight, these are the forces acting on motor when it moves in a straight line

Answer:

its temperature must increase.

Explanation:

An isobaric expansion is a transformation in which the pressure of the gas does not change (isobaric), while the volume increases (expansion).

Since the pressure does not change,

"its pressure must increase."

is a false statement.

The 1st law of thermodynamics is

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

where

[tex]\Delta U[/tex] is the change in internal energy of the gas, which is proportional to the change in temperature: [tex]\Delta U \propto \Delta T[/tex]

Q is the heat supplied to the gas

W is the work done by the gas, which is given by

[tex]W=p\Delta V[/tex]

where p is the pressure and [tex]\Delta V[/tex] is the change in volume. Since the gas is expanding, we can say that [tex]\Delta V>0[/tex], so the gas does positive work:

[tex]W>0[/tex]

This means that the option

"the gas does no work."

is false.

Moreover, from the ideal gas law

[tex]pV=nRT[/tex] (2)

we also know that the temperature of the gas is increasing (because p, the pressure, n the number of moles, and R, the gas constant, are all constant in this process, and since the volume V is increasing, than the temperature T must be increasing also)

So, we know that the option

"its internal (thermal) energy does not change. "

is false.

Finally, in an isobaric expansion, in order to keep the pressure constant heat should be supplied to the system, so

"no heat enters or leaves the gas."

is also wrong

We also said from (2) that the temperature of the gas is increasing, therefore the statement

"its temperature must increase."

is the only correct one.

Answer:

F. The mesosphere contains fewer oxygen molecules than the stratosphere.

Explanation:

The layers of the atmosphere are divided into:

1. Troposphere

2. Stratosphere

3. Mesosphere

4. Thermosphere

5. Exosphere

The troposphere extends from the earth surface to about 10km upwards. This is the region of the greatest atmospheric pressure and where all weather conditions arises. In the troposphere, the higher one goes the cooler it becomes.

The stratosphere lies on the troposphere and it is about 50km from the top of the troposphere. The stratosphere is the region where ozone, an oxygen molecule, forms a layer. The higher you go in the stratosphere, the warmer it becomes.

The mesosphere is about 90km thick and it has less gas density. It extends from the top of the stratosphere upwards. The gases here are sparse and atmospheric pressure is lesser than that of the surface. Here, the higher you go in the mesosphere, the cooler it becomes.

Only option F is correct: the mesosphere contains fewer oxygen molecules than the stratosphere.

Answer:

F: The mososphere contains fewer oxygen molecules than the Stratosphere.

Explanation:

Answer:

754.6 m

Explanation:

The GPE (Gravitational potential energy) of an object with respect to the ground is given by

[tex]GPE = mgh[/tex]

where

m is the mass of the object

g = 9.8 m/s^2 is the acceleration due to gravity

h is the heigth above the ground

Here we have

m = 12,400.05 kg is the mass

GPE = 91,700,000.00J is the GPE

Solving the formula for h, we find the heigth:

[tex]h=\frac{GPE}{mg}=\frac{91,700,000.00J}{(12,400.05 kg)(9.8 m/s^2)}=754.6 m[/tex]

Final answer:

The 12,400.05kg generator with a gravitational potential energy of 91,700,000J was approximately 750 meters above the ground, calculated using the formula for GPE which is GPE = mgh.

Explanation:

The student asked how far above the ground a 12,400.05kg generator was when it had a gravitational potential energy (GPE) of 91,700,000.00J. To find the height, we use the formula for gravitational potential energy: GPE = mgh, where m is mass in kilograms, g is the acceleration due to gravity (9.8 m/s2), and h is the height in meters. Solving for h, we rearrange the formula to h = GPE / (mg).

By inserting the values we have:

h = 91,700,000J / (12,400.05kg × 9.8m/s2)

After calculating:

h ≈ 750 meters

Therefore, the generator was approximately 750 meters above the ground.

Answer:

[tex]2.25\cdot 10^5 J[/tex]

Explanation:

First of all, we have to find the power used by the hair dryer, which is given by

[tex]P=\frac{V^2}{R}[/tex]

where

V = 120 V is the voltage

[tex]R=9.6 \Omega[/tex] is the resistance of the hair dryer

Substituting,

[tex]P=\frac{(120 V)^2}{9.6 \Omega}=1500 W[/tex]

Now we can find the total electrical energy used, given by

[tex]E=Pt[/tex]

where P is the power and

t = 2.5 min = 150 s is the time

Substituting,

[tex]E=(1500 W)(150 s)=2.25\cdot 10^5 J[/tex]

Answer:

4 I

Explanation:

The initial resistance of the wire is given by:

[tex]R=\frac{\rho L}{A}[/tex]

where

[tex]\rho[/tex] is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

Since the resistance is proportional to the length of the wire, when the wire is cut in half, each wire will have half of the initial resistance:

[tex]R' = \frac{\rho \frac{L}{2}}{A}=\frac{R}{2}[/tex]

Later, these two pieces of wire are connected in parallel to the battery (because the right ends of both wires are attached to one terminal, while the left ends are attached to the other terminal, so they are connected in parallel). Therefore, their total resistance will be:

[tex]\frac{1}{R_T} = \frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R/2}+\frac{1}{R/2}=\frac{2}{R}+\frac{2}{R}=\frac{4}{R}\\R_T = \frac{R}{4}[/tex]

so the total resistance of the circuit will be 1/4 of the initial resistance.

The initial current in the circuit was given by Ohm's law:

[tex]I=\frac{V}{R}[/tex]

Since the voltage V has not changed, the new resistance will be

[tex]I' = \frac{V}{R_T}=\frac{V}{R/4}=4 \frac{V}{R}=4I[/tex]

so, the current will increase by a factor 4.

When the length of the wire is halved and connected in parallel, then the current increases four times the initial current.

Ohm's law states that when there is no physical change in the conductor, the potential difference is directly proportional to the current flowing in it.

V = RI

V be the potential difference

I be the current.

R be the resistance of the circuit.

A resistor is made out of a wire having a length of L.

When the ends of the wire are attached across the terminals of an ideal battery having a constant voltage V₀ across its terminals, a current I flows through the wire.

We know that the resistance is given by

[tex]\rm R = \dfrac{\rho L}{A}[/tex]

where L is the length, A is the area, and ρ is the resistivity of the conductor.

If the wire were cut in half, making two wires of length L/2, and both wires were attached across the terminals of the battery. Then we have

[tex]\rm R '= \dfrac{\rho \frac{L}{2}}{A} = \dfrac{R}{2}[/tex]

These two wires are in parallel. Then we have

[tex]\rm \dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_1}\\\\\\\dfrac{1}{R_{eq}} = \dfrac{1}{\frac{R}{2}}+\dfrac{1}{\frac{R}{2}}\\\\\\R_{eq } \ = \dfrac{R}{4}[/tex]

Then the current will be given by Ohm's law

[tex]\rm I' = \dfrac{V}{R_{eq}}\\\\\\I' = \dfrac{V}{\frac{R}{4}} \\\\\\I' = 4 \dfrac{V}{R} \\\\\\ I ' = 4 \ I[/tex]

More about Ohm's law link is given below.

https://brainly.com/question/796939

Answer:

d = 84 m

Explanation:

As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object

here we know that

initial speed [tex]v_i = 42 m/s[/tex]

final speed [tex]v_f = 0[/tex]

time taken by the car to stop

[tex]t = 4s[/tex]

now the distance moved by the car before it stop is given as

[tex]d = \frac{v_f + v_i}{2} \times t [/tex]

now we have

[tex]d = \frac{42 + 0}{2} \times 4[/tex]

[tex]d = 84 m[/tex]

The car that travels to the right with a speed of 42 m/s, skids 84 meters for 4 seconds before it comes to a stop.

The distance traveled by car before coming to a stop can be calculated with the following equation:

[tex] v_{f}^{2} = v_{i}^{2} + 2ad [/tex]    (1)

Where:

[tex] v_{f}[/tex]: is the final speed = 0 (it stops)

[tex] v_{i}[/tex]: is the initial speed = 42 m/s

a: is the acceleration

d: is the distance =?  

We need to find the acceleration. We can use the next equation:

[tex] v_{f} = v_{i} + at [/tex]    (2)  

Where:

t: is the time = 4.0 s

Hence, the acceleration is:

[tex]a = \frac{v_{f} - v_{i}}{t} = \frac{0 - 42 m/s}{4.0 s} = -10.5 m/s^{2}[/tex]

Now, the car skid the following meters before coming to a stop (eq 1).

[tex]d = \frac{v_{f}^{2} - v_{i}^{2}}{2a} = \frac{-(42 m/s)^{2}}{2(-10.5 m/s^{2})} = 84 m[/tex]  

Therefore, the car skids 84 meters before coming to a stop.

To find more about stopping distance, go here: https://brainly.com/question/4299689?referrer=searchResults

I hope it helps you!      

Answer:MA = 15

Explanation:The mechanical advantage for an inclined plane is MA=l/h or length divided by height. So, plugging these variables into the equation would have it set up like this: MA = 30/2. When 30 is divided by 2 you get your answer for mechanical advantage, which would be 15

Answer:

the answer is 15

Explanation:

i got it right

Answer:

Transverse wave

Explanation:

Depending on the direction of the oscillation, there are two types of waves:

- Transverse wave: in a transverse wave, the oscillation occurs in a direction perpendicular to the direction of propagation of the wave. These types of waves are characterized by alternating crests and troughs. Examples of transverse waves are electromagnetic waves.

- Longitudinal wave: in a longitudinal wave, the oscillation occurs in a direction parallel to the direction of propagation of the wave. These types of waves are characterized by alternating regions of higher density (compressions) and lower density (rarefactions). Examples of longitudinal waves are sound waves.

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.

Answer:

1. The current will drop to half of its original value.

Explanation:

The problem can be solved by using Ohm's law:

[tex]V=RI[/tex]

where

V is the voltage across the circuit

R is the resistance of the circuit

I is the current

We can rewrite it as

[tex]I=\frac{V}{R}[/tex]

In this problem, we have:

- the resistance of the circuit remains the same: R' = R

- the voltage is decreased to half of its original value: [tex]V'=\frac{V}{2}[/tex]

So, the new current will be

[tex]I'=\frac{V'}{R'}=\frac{V/2}{R}=\frac{1}{2}\frac{V}{R}=\frac{I}{2}[/tex]

so, the current will drop to half of its original value.

Final answer:

If the voltage across a circuit decreases to half its original value while the resistance remains constant, the current will drop to half of its original value.

Explanation:

According to Ohm's law, the current in a circuit is directly proportional to the voltage and inversely proportional to the resistance. If the voltage across a circuit decreases to half its original value while the resistance remains constant, the current will also decrease proportionally. Therefore, the correct answer is option 1: The current will drop to half of its original value.

Answer:

These are basically all neon gasses

Explanation:

An example can be argon

The energy that is put into the system is conserved and transferred to the water to increase its internal energy, which causes it to boil. The electric energy is converted to heat energy which is inturn converted to kinetic energy for the water to boil and form steam.

When a pan of water is placed on an electric stove and heated, energy is transferred from the stove to the water through conduction. The electric stove uses electricity to heat up the heating element, which then transfers the heat to the bottom of the pan via conduction.

As the water in the pan heats up, the molecules in the water gain energy and begin to move faster. Eventually, the water reaches its boiling point, which is when the molecules have enough energy to break the bonds between them and turn into steam.

During this process, energy is conserved because energy cannot be created or destroyed, only transferred or converted from one form to another. In this case, the electrical energy from the stove is converted into thermal energy, which is then transferred to the water through conduction. The energy that is transferred to the water is then used to increase the kinetic energy of the water molecules and eventually to break the bonds between the molecules to form steam.

In other words, the energy that is put into the system is conserved and transferred to the water to increase its internal energy, which causes it to boil. This process will continue until all the water in the pan has turned into steam, at which point the energy that was put into the system will be conserved in the form of the steam's increased kinetic energy.

Learn more on the conversion of energy here https://brainly.com/question/961052

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Answer:

7.1 Hz

Explanation:

In a generator, the maximum induced emf is given by

[tex]\epsilon= 2\pi NAB f[/tex]

where

N is the number of turns in the coil

A is the area of the coil

B is the magnetic field strength

f is the frequency

In this problem, we have

N = 200

[tex]A=0.030 m^2[/tex]

[tex]\epsilon=8.0 V[/tex]

B = 0.030 T

So we can re-arrange the equation to find the frequency of the generator:

[tex]f=\frac{\epsilon}{2\pi NAB}=\frac{8.0 V}{2\pi (200)(0.030 m^2)(0.030 T)}=7.1 Hz[/tex]

The frequency of the generator is about 7.1 Hz

[tex]\texttt{ }[/tex]

Let's recall eletromotive force formula for a generator:

[tex]\large {\boxed {\varepsilon = NBA\omega \sin (\omega t) }[/tex]

ε = electromotive force ( V )

B = magnetic field strength (T)

N= number of turns in a coil

A = cross-sectional area ( m² )

ω = angular frequency ( rad/s )

t = time taken ( s )

Let's tackle the problem now !

[tex]\texttt{ }[/tex]

Given:

maximum emf = ε = 8.0 V

number of turns = N = 200 turns

cross-sectional area = A = 0.030 m²

magnetic field strength = B = 0.030 T

Asked:

frequency of the generator = f = ?

Solution:

[tex]\varepsilon = NBA\omega \sin (\omega t)[/tex]

[tex]\varepsilon_{max} = NBA\omega[/tex]

[tex]\varepsilon_{max} = NBA(2 \pi f)[/tex]

[tex]f = \varepsilon_{max} \div ( 2\pi NBA )[/tex]

[tex]f = 8.0 \div [ 2\pi (200)(0.030)(0.030) ][/tex]

[tex]f = 8.0 \div [ 0.36 \pi ][/tex]

[tex]f = \frac{200}{9 \pi}[/tex]

[tex]f \approx 7.1 \texttt{ Hz}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Magnetic Field

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Keywords: Magnet , Field , Magnetic , Current , Wire , Unit ,