A cheetah can run at 105 feet per second, but only for 7 seconds, at which time the animal must stop and rest. A fully rested cheetah at (0, 0) notices a nearby antelope, which is moving according to the parametric equation (x, y) = (−39 + 40t, 228 + 30t), where t is measured in seconds and x and y are measured in feet. If it started to run at t = 0, the cheetah could catch the antelope. For how many more seconds can the cheetah afford to wait before starting? Assume that the cheetah does not change direction when it runs.

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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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.

20.november1998 im pretty sure

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.

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]

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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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: 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.

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:

conversion of mass into energy

Explanation:

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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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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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.

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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I think the answer is d

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.

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:

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

[tex]\texttt{ }[/tex]

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

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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I'm pretty that the answer would be D.

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:

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]

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

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:

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.

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]

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:

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.  

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.