Match the descriptions to the feature,
principal axis
Center of the spherical mirror from which a
curved mirror was cut
center of curvature
Distance from the center of a mirror to the
focal point
Line that runs through the center of
curvature to a mirror
focal point
Point where reflected light converges or
appears to diverge
focal length
Point where the principal axis and mirror
vertex
meet

Match The Descriptions To The Feature,principal AxisCenter Of The Spherical Mirror From Which Acurved

Answers

Answer 1

Answer:

Vertex: Point where the principal axis and mirror meet

Focal point: Point we are reflected light converges or appears to diverge

Focal length: distance from the center of a mirror to the focal point

Principal axis: line that runs to the center of curvature to a mirror

Center of curvature: sensor of spherical mirror from which a curved mirror was cut

Explanation:

Just did the assignment on Edge.

Match The Descriptions To The Feature,principal AxisCenter Of The Spherical Mirror From Which Acurved
Answer 2

A spherical mirror is a mirror which has the shape of a piece cut out of a spherical surface. There are two types of spherical mirrors: concave, and convex.

What is a spherical mirror?

A spherical mirror is a mirror which has the shape of a piece cut out of a spherical surface. There are two types of spherical mirrors: concave, and convex. These are illustrated in.

The most commonly occurring examples of concave mirrors are shaving mirrors and makeup mirrors.

As is well-known, these types of mirrors magnify objects placed close to them. The most commonly occurring examples of convex mirrors are the passenger-side wing mirrors of cars.

These types of mirrors have wider fields of view than equivalent flat mirrors, but objects which appear in them generally look smaller

The answers to the given questions are:

Vertex: Point where the principal axis and mirror meet

Focal point: Point we are reflected light converges or appears to diverge

Focal length: distance from the centre of a mirror to the focal point

Principal axis: line that runs to the centre of curvature of a mirror

Center of curvature: sensor of spherical mirror from which a curved mirror was cut

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Related Questions

A cart of mass 300 g is placed on a frictionless horizontal air track. A spring having a spring constant of 9.0 N/m is attached between the cart and the left end of the track. The cart is displaced 3.8 cm from its equilibrium position. (a) Find the period at which it oscillates. Correct: Your answer is correct. s (b) Find its maximum speed. Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s (c) Find its speed when it is located 2.0 cm from its equilibrium position. m/s

Answers

Answer:

Explanation:

the angular frequency ω    of the pendulum is given by the formula

ω  = [tex]\sqrt{\frac{k}{m} }[/tex]    , k is spring constant , m is mass attached .

= [tex]\sqrt{\frac{9}{.3} }[/tex]

= 5.48 rad /s

time period = 2π / ω

= 2 x 3.14 / 5.48

= 1.146 s

b )  formula for speed

v = ω[tex]\sqrt{(a^2-\ x^2)}[/tex]    , a is amplitude , x is displacement from equilibrium point.

for maximum speed x = 0

max speed = ωa

= 5.48 x 3.8 x 10⁻²  ( initial displacement becomes amplitude that is 3.8 cm )

= .208 m /s

20.8 cm / s

c )

when x = .02 m , velocity = ?

v = ω[tex]\sqrt{(a^2-\ x^2)}[/tex]  

=  5.48 [tex]\sqrt{(.038^2-\ .02^2)}[/tex]

= 5.48 x .0323109

= .177 m /s

17.7 cm /s .

When a rocket is traveling toward a mountain at 100 m/s, the sound waves from this rocket's engine approach the mountain at speed V.

If the rocket doubles its speed to 200 m/s, the sound waves from the engine will now approach the mountain at speed

A. 2–√V
B. 2V
C. V
D. 4V

Answers

Answer:

The correct option is C

Explanation:

From the question we are told that

  The initial speed of the rocket is  [tex]v_i = 100 m/s[/tex]

    The speed of the rocket engine sound is [tex]V[/tex]

    The final speed of the rocket is  [tex]v_f = 200 \ m/s[/tex]

The speed of the sound at [tex]v_f[/tex] would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).

  A clear example when  lightning strikes you will first see (that is because it travels at  the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the   lightning

Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates  

A uniform-density 8 kg disk of radius 0.25 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 41 N through a distance of 0.9 m. Now what is the angular speed

Answers

Answer:

Explanation:

moment of inertia of the disc I = 1/2 m r²

= .5 x 8 x .25²

= .25 kg m²

The wok done by force will be converted into rotational kinetic energy

F x d = 1/2 I ω²

F is force applied , d is displacement , I is moment of inertia of disc and ω

is angular velocity of disc

41 x .9 = 1/2 x .25 ω²

ω² = .25

ω = 17.18  rad / s

The angular speed should be 17.18  rad / s

Calculation of the angular speed:

Since

moment of inertia of the disc I = 1/2 m r²

= .5 x 8 x .25²

= .25 kg m²

Now the work done by force should be converted into the rotational kinetic energy

F x d = 1/2 I ω²

here,

F is the force applied,

d is displacement,

I is moment of inertia of disc

and ω is angular velocity of disc

So,

41 x .9 = 1/2 x .25 ω²

ω² = .25

ω = 17.18  rad / s

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For an experiment, a student wants to maximize the power output for a circuit. The student has three resistors of resistances 512, 1012, and 20 S, and a 5 V battery. Which of the following best describes how the students should arrange the resistors so that when they are connected to the battery it will maximize the power output for the circuit and explains why? a) The student should place all resistors in parallel, because the battery supplies more current to a parallel combination than to a series combination b) The student should place all resistors in parallel, because the potential difference across all the resistors is less for a parallel combination than for a series combination c) The student should place all resistors in series, because this creates the largest equivalent resistance d) The student should place all resistors in series, because the battery supplies more current to a series combination than to a parallel combination e) The student should place all resistors in series, because the potential difference across all the resistors is greater in series than in parallel

Answers

Final answer:

To maximize the power output of a circuit with a 5V battery and three resistors, the resistors should be placed in parallel because this arrangement supplies more current than a series combination, thereby maximizing power.

Explanation:

To maximize the power output for a circuit with a given voltage source, the resistors should be arranged to draw the highest current possible. For resistors connected in parallel, the total resistance of the circuit decreases, as opposed to when they are in series, where it increases. Since power (P) is calculated using the formula P = V^2 / R, where V is the voltage and R is the resistance, minimizing R will maximize P.

Because a parallel configuration guarantees that each resistor gets the full voltage of the battery, the correct answer is: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination. This is because in a parallel circuit, the voltage across each resistor is equal to the voltage of the source, which maximizes the current through each resistor due to Ohm's law (I = V/R), and therefore maximizes the total current in the circuit.

Answer option (a) is thus correct: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination.

Follow these steps to solve this problem: Two identical loudspeakers, speaker 1 and speaker 2, are 2.0 m apart and are emitting 1700-Hz sound waves into a room where the speed of sound is 340 m/s. Consider a point 4.0 m in front of speaker 1, which lies along a line from speaker 1, that is perpendicular to a line between the two speakers. Is this a point of maximum constructive interference, a point of perfect destructive interference, or something in between?

Answers

Answer:

It is somewhere in between

Explanation:

Wave length of sound from each of the  speakers = 340 / 1700 = 0.2 m = 20 cm

Distance between first speaker and the given point = 4 m.

Distance between second speaker and the given sound

D = √(4² + 2²)

D = √(16 + 4)

D = √20

D = 4.472 m

Path difference = 4.472 - 4 =

0.4722 m.

Path difference / wave length = 0.4772 / 0.2 = 2.386

This is a fractional integer which is neither an odd nor an  even multiple of half wavelength. Hence this point is of neither a perfect constructive nor a perfect destructive interference.

Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newton's Law of Cooling for this particular situation. (Use t as the independent variable, y as the dependent variable, R as the room temperature, and k as a proportionality constant.)

Answers

Answer:

[tex]\frac{dy}{dt}=k(y_t-R)[/tex]

Explanation:

According to Newton’s law of cooling, the rate of loss of heat from a body and the difference in the temperature of the body and its surroundings are proportional to each other.

[tex]\frac{dy}{dt}=k(y_t-R)[/tex]

Here, [tex]y_t[/tex] represents temperature at time t, R as the room temperature, t as the independent variable, y as the dependent variable.

The equation that represents Newton's Law of Cooling for this particular situation is dy/dt = k(yt-R)

What is Newton’s law?

According to Newton’s law of cooling refer, the rate of loss of heat from a body and also the difference in the temperature of the body and also its surroundings are proportional to each other.

dy/dt is = k(yt-R)

Therefore, yt conveys temperature at time t, R as the room temperature, t as the independent variable, y as the dependent variable.

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A girl launches a toy rocket from the ground. The engine experiences an average thrust of 5.26 N. The mass of the engine plus fuel before liftoff is 25.0 g, which includes fuel mass of 12.7 g. The engine fires for a total of 1.90 s. (Assume all the fuel is consumed.) (a) Calculate the average exhaust speed of the engine (in m/s). (b) This engine is situated in a rocket case of mass 63.0 g. What is the magnitude of the final velocity of the rocket (in m/s) if it were to be fired from rest in outer space with the same amount of fuel

Answers

Answer:

a) v = 786.93 m/s

b) v = 122.40 m/s

Explanation:

a) To find the average exhaust speed (v) of the engine we can use the following equation:

[tex] F = \frac{v\Delta m}{\Delta t} [/tex]

Where:

F: is the thrust by the engine = 5.26 N

Δm: is the mass of the fuel = 12.7 g

Δt: is the time of the burning of fuel = 1.90 s

[tex]v = \frac{F*\Delta t}{\Delta m} = \frac{5.26 N*1.90 s}{12.7 \cdot 10^{-3} kg} = 786.93 m/s[/tex]

b) To calculate the final velocity of the rocket we need to find the acceleration.

The acceleration (a) can be calculated as follows:

[tex] a = \frac{F}{m} [/tex]

In the above equation, m is an average between the mass of the engine plus the rocket case mass and the mass of the engine plus the rocket case minus the fuel mass:

[tex]m = \frac{(m_{engine} + m_{rocket}) + (m_{engine} + m_{rocket} - m_{fuel})}{2} = \frac{2*m_{engine} + 2*m_{rocket} - m_{fuel}}{2} = \frac{2*25.0 g + 2*63.0 g - 12.7 g}{2} = 81.65 g[/tex]

Now, the acceleration is:

[tex] a = \frac{5.26 N}{81.65 \cdot 10^{-3} kg} = 64.42 m*s^{-2} [/tex]

Finally, the final velocity of the rocket can be calculated using the following kinematic equation:

[tex]v_{f} = v_{0} + at = 0 + 64.42 m*s^{-2}*1.90 s = 122.40 m/s[/tex]

I hope it helps you!

Final answer:

The average exhaust speed of the engine is calculated to be approximately 786.929 m/s, and the magnitude of the final velocity of the rocket if fired in outer space is roughly 94.916 m/s.

Explanation:

To calculate the average exhaust speed (v_ex), we can use the impulse-momentum theorem, which states that impulse is equal to the change in momentum of the system. Impulse is given by the product of the average force exerted by the engine (thrust) and the time interval during which the thrust is applied. If all the fuel is consumed, the change in mass (Δm) is the mass of the fuel.

Impulse = Thrust × Time = 5.26 N × 1.90 s = 9.994 N·s

The momentum change is equal to the mass of the fuel expelled times the average exhaust speed.

Δ(momentum) = Δm × v_ex

Substituting the impulse and solving for v_ex, we get:

v_ex = Impulse / Δm

v_ex = 9.994 N·s / 0.0127 kg = 786.929 m/s

Part B: Final Velocity of the Rocket in Space

The final velocity (V_final) of the rocket in space can be determined using the rocket equation, also known as Tsiolkovsky's rocket equation:

V_final = v_ex × ln(m_initial / m_final)

Where:

m_initial = mass of the rocket with fuel = 25.0 g + 63.0 g = 88.0 g = 0.088 kgm_final = mass of the rocket without fuel = 25.0 g - 12.7 g + 63.0 g = 75.3 g = 0.0753 kg

Calculating the final velocity:

V_final = 786.929 m/s × ln(0.088 kg / 0.0753 kg) ≈ 94.916 m/s

One particle has a mass of 3.12 x 10-3 kg and a charge of +8.8 C. A second particle has a mass of 7.1 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.15 m, the speed of the 3.12 x 10-3 kg particle is 131 m/s. Find the initial separation between the particles.

Answers

Answer:[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]

Explanation:

Given

mass of first particle is [tex]m_1=3.12\times 10^{-3}\ kg[/tex]

mass of second particle is [tex]m_2=7.1\times 10^{-3}\ kg[/tex]

Charge on both the particle [tex]q=8.8\times 10^{-6}\ C[/tex]

Now final speed of first particle is [tex]v_1=131\ m/s[/tex]

Final separation between particles is [tex]r=0.15\ m[/tex]

As there is no external force therefore linear momentum is conserved

[tex]0+0=m_1v_1+m_2v_2[/tex]

[tex]0=3.12\times 10^{-3}\times 131+7.1\times 10^{-3}\times v_2[/tex]

[tex]v_2=-\dfrac{3.12\times 10^{-3}}{7.1\times 10^{-3}}\times 131[/tex]

[tex]v_2=-57.56\ m/s[/tex]

Conserving total energy

Initial Kinetic energy +Initial  Potential energy=Final Kinetic energy +Final Potential energy

[tex]\Rightarrow 0+\frac{kq^2}{r_i}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{kq^2}{r_f}[/tex]

[tex]\Rightarrow \frac{9\times 10^9\times 8.8^2\times 10^{-12}}{r_i}=\frac{1}{2}\times 3.12\times 10^{-3}\times 131^2+\frac{1}{2}7.1\times 10^{-3}\times (57.56)^2+\frac{9\times 10^9\times 8.8^2\times 10^{-12}}{0.15}[/tex]

[tex]\Rightarrow \frac{0.696}{r_i}=26.771+11.76+4.646[/tex]

[tex]\Rightarrow \frac{0.696}{r_i}=43.177[/tex]

[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]

Using the law of conservation of linear momentum and conservation of total energy, the obtained initial separation between the particles is 0.01612 m.

Conservation of Linear Momentum

Given that the masses of the two particles are;

[tex]m_1 = 3.12\times 10^{-3}\,kg\\m_2 = 7.1\times 10^{-3}\,kg[/tex]

Also, the charges of both the particles are equal.

[tex]q_1 = q_2 = +8.8\times 10^{-6} \,C[/tex]

The final separation between the particles is;

[tex]r_f = 0.15\,m[/tex]

Also, the final speed of the first particle is;

[tex](v_1)_{final} = 131\,m/s[/tex]

There is no external force applied here; so by the law of conservation of linear momentum, we have;

[tex](m_1v_1)_{initial} +(m_2v_2)_{initial}=(m_1v_1)_{final} +(m_2v_2)_{final}[/tex]

But initially, the particles are at rest, so the initial velocities are zero.

[tex]0+0=(3.12\times 10^{-3}\,kg \, \times 131\,m/s ) +(7.1\times 10^{-3}\,kg\,)\,(v_2)_{final}\\[/tex]

[tex]\implies (v_2)_{final}=-\frac{3.12\times 10^{-3}\,kg \, \times 131\,m/s}{7.1\times 10^{-3}\,kg} =-57.57\,m/s[/tex]

Conservation of Energy

Now, applying the law of conservation of total energy, we get;

[tex](KE)_{\,initial}\, + \,(PE)_{\,initial} = (KE)_{\,final}\, + \,(PE)_{\,final}[/tex]

But initially, the particles are at rest; so they have no initial kinetic energy.

They have electrostatic potential alone initially.

[tex]0+k\frac{q^2}{r_i} = \frac{1}{2}(m_1 v_1)_{initial} + \frac{1}{2}(m_2 v_2)_{final} + k\frac{q^2}{r_f}[/tex]

Substituting the known values, we get;

[tex](9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_i} =[ \frac{1}{2}(3.12\times 10^{-3}\,kg )\times (131\,m/s)^2] + [\frac{1}{2}(7.1\times 10^{-3}\,kg) \times (-57.57\,m/s)^2 + (9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_f}[/tex]

[tex]\implies\frac{0.696\,Nm^2}{r_i} =26.77\,J+11.76\,J+ 4.64\,J[/tex]

[tex]\implies r_i =\frac{0.696\,Nm^2}{43.17\,J}=0.01612\,m[/tex]

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Describe the process of nuclear fission. How does it work and when is it used? What is meant by a chain reaction, and how do you control it?

Answers

Nuclear fission is a nuclear reaction in which a heavy nucleus splits spontaneously or on impact with another particle, with the release of energy. And is used to spin a turbine to produce electricity.

Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+β .

Answers

Final answer:

To evaluate the final kinetic energy of a supply spacecraft under the given tractor beam force, you would need to integrate the force over the displacement. Direct calculation requires specific values for α, β, and displacement x, which are not provided.

Explanation:

The question involves calculating the final kinetic energy of a supply spacecraft under a specific tractor beam force, F(x)=αx3+β. This calculation would typically require the integration of the force over the displacement, as kinetic energy can be evaluated through the work-energy principle where work done by a force in moving an object is equal to the change in kinetic energy of the object. Here, without specific values or further context provided for α, β, or the displacement, x, a direct calculation cannot be made. However, in physics, especially in the study of mechanics and spacecraft dynamics, understanding how forces affect motion and energy forms the basis for analyzing and optimizing space missions. Evaluating the final kinetic energy would typically involve integrating the force function over the spacecraft's path, considering initial conditions, and any external forces or resistances.

The potential energy of a 1.7 x 10-3 kg particle is described by U (x )space equals space minus (17 space J )space cos open square brackets fraction numerator x over denominator 0.35 space straight m end fraction close square brackets. What is the angular frequency of small oscillations around the point x = 0?

Answers

Answer:

Explanation:

Given a particle of mass

M = 1.7 × 10^-3 kg

Given a potential as a function of x

U(x) = -17 J Cos[x/0.35 m]

U(x) = -17 Cos(x/0.35)

Angular frequency at x = 0

Let find the force at x = 0

F = dU/dx

F = -17 × -Sin(x/0.35) / 0.35

F = 48.57 Sin(x/0.35)

At x = 0

Sin(0) =0

Then,

F = 0 N

So, from hooke's law

F = -kx

Then,

0 = -kx

This shows that k = 0

Then, angular frequency can be calculated using

ω = √(k/m)

So, since k = 0 at x = 0

Then,

ω = √0/m

ω = √0

ω = 0 rad/s

So, the angular frequency is 0 rad/s

En una competición de tiro con arco, la diana de 80 cm de diámetro se encuentra a 50 m de distancia y a 1,5 metros del suelo. En uno de los tiros la flecha sale a 230 km hora con una de 3,5 grados desde una altura de 1,60 metros,despreciando el rozamiento con el aire.¿ Impactará la flecha en la diana? En caso afirmativo¿ con qué velocidad y en qué dirección?

Answers

Answer:

a .

15.68

m/s

(b).  

21.72

m

Explanation:

To determine if the arrow will hit the target, we can analyze the projectile motion. By calculating the time of flight and horizontal displacement, we can determine if the arrow will hit the target. The velocity and direction of the arrow at impact can also be determined using the horizontal and vertical components of velocity.

To determine if the arrow will hit the target, we need to analyze the projectile motion of the arrow. First, we need to calculate the time of flight using the equation t = 2 * (v * sin(theta))/g, where v is the initial velocity of the arrow and theta is the launch angle. Next, we can calculate the horizontal displacement using the equation x = v * cos(theta) * t, where x is the distance traveled horizontally. Comparing the calculated horizontal displacement with the distance to the target, we can determine if the arrow will hit the target.

Given the initial velocity of the arrow (230 km/hr), launch angle (3.5 degrees), and distance to the target (50 m), we can convert the initial velocity to m/s (230 km/hr * (1000 m/km) * (1 hr/3600 s)) and use the formulas to calculate time of flight and horizontal displacement. If the horizontal displacement is less than or equal to the distance to the target, the arrow will hit the target.

To find the velocity and direction of the arrow at impact, we can use the equations v_x = v * cos(theta) and v_y = v * sin(theta), where v_x is the horizontal component of velocity and v_y is the vertical component of velocity. Since the angle of impact is not specified, the direction of the arrow at impact will depend on the particular situation.

Does this chemical equation support the Law of Conservation of Mass? Why or why not?
N₂ + H₂ → NH₃

Answers

Answer:No it does not the chemical equation is unbalanced,this is because according to the law of conservation it states that mass cannot be created or destroyed ,the chemical equation N₂ + H₂ → NH₃ shows that energy has been created which is not possibly so the correct balanced chemical is equation is  1N₂ +3H₂ → 3NH₃

Explanation:

Use the drop-down menus to complete the scenarios.
A patient has an ongoing history of cancer. She has a tumor in the abdominal region, and has been undergoing
treatment for it. There may be other tumors and a potential blockage in the surrounding area that need to be
investigated. The imaging technique that might provide the most information in this case is
Joe has ongoing issues with his throat and feels some sort of blockage or abnormality as he swallows. The doctor
decides to use X-ray imaging to visualize Joe's internal anatomy as he swallows to help determine the nature of the
problem.
will be used for this procedure.

Answers

First: CT Scan

Second: Fluoroscopy

Explanation:

Both correct

Final answer:

A CT scan can provide detailed images of a tumor and its surrounding area, which can be beneficial for a cancer patient. For ongoing throat issues, a Barium swallow study can be used, where the patient swallows a barium solution that is then visualized with an X-ray to identify abnormalities.

Explanation:

In the first scenario, the patient with an ongoing history of cancer might benefit most from a Computed Tomography (CT) scan. CT scans are capable of creating detailed pictures of organs, bones, and other tissues, making it an excellent tool for capturing the size and position of a tumor and surrounding blockages in the abdominal area. Furthermore, it can reveal whether the cancer has spread to other parts of the body.

In the second scenario, Joe's doctor chooses to use X-ray imaging to detect any abnormalities in his throat. The specific procedure used for this is known as a Barium swallow study. This involves swallowing a barium solution that coats the esophagus, enabling the X-ray to capture clear images of the region as the patient swallows.

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A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for gas heating and cooking in a period of 1 month. Given that 1.00 therm is equal to 29.329.3 kWh, how many milligrams of mass would need to be converted directly to energy each month to meet the energy needs for the home?

Answers

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

Final answer:

Total monthly energy consumption of the house in kilowatt-hours (kWh) is calculated. Then, with the help of the equation E = mc², where c is the speed of light, the equivalent mass of energy in kilograms is found by converting energy to Joules and solving for mass.

Explanation:

The household uses 475475 kWh of electrical energy and 135135 therms for gas heating and cooking monthly. Considering that 1.00 therm is equal to 29.329.3 kWh, total energy consumption for the month in kWh can be calculated as (475475 + (135135 × 29.3)) kWh.

Einstein’s famous equation E = mc² relating energy (E) and mass (m) allows us to calculate the equivalent mass of this energy.

Here, c is the speed of light. We convert the energy into Joules (1 kWh = 3.6 × 10⁶ J) and then solve for m to find out the mass in kilograms that would need to be converted into energy each month. The question seems to have a typo in the values, you might need to correct them to calculate the accurate amount.

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When you run around a track at 5 km/h, your velocity is constant. Please select the best answer from the choices provided T F

Answers

Answer:

T

Explanation:

Answer:

true

Explanation:

Write a function to model the volume of a rectangular prism if the length is 24cm and the sum of the width and height is 28cm

Answers

[tex]w + h = 28[/tex]

[tex]l \times w \times h = v[/tex]

[tex]v = w \times h \times 24[/tex]

[tex]w = 28 - h[/tex]

[tex]h(28 - h) \times 24 = v[/tex]

[tex]24( - {h}^{2} + 28h) = v[/tex]

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.1 cm ( 0.581 m) and the flow speed of the petroleum is 10.1 m/s. At the refinery, the petroleum flows at 5.85 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?


volume flow rate:________m^3/sdiameter________cm

Answers

Answer:

volume flow rate:2.68m³/s

diameter :58.27cm

Explanation:

flow rate = Q = πr² v  = amount per second that flows through the pipe

Given:

pipe's diameter= 0.581

r= 0.581/2=>0.2905m

speed 'v'= 10.1 m/s

Q= (3.142)(0.2905)²(10.1)

so

volume flow rate= 2.68m³/s

->if no oil has been added or subtracted or compressed then Q is the same everywhere

therefore,

Q= πr²v

2.68 = π(d/2)²(5.85)

d= (2.68x4) /(3.142 x 5.85)

d= 0.5827m =>58.27cm

A 0.5 Kg pinball is initially at rest against a 120 N/m spring. The shooter is pulled back and has the spring compressed a distance of 0.2 m. The spring is released and the ball is shot up the ramp. It hits nothing and eventually comes to rest before it begins to roll down. We can ignore friction. The game board ramp is at an angle of 30o . How far did the ball travel on the board from the place of maximal compression to the first stop

Answers

Answer:

[tex]\Delta s = 0.978\,m[/tex]

Explanation:

The pinball-spring system is modelled after the Principle of Energy Conservation:

[tex]U_{g,1} + U_{k,1} + K_{1} = U_{g,2} + U_{k,2} + K_{2}[/tex]

[tex]-(0.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right) \cdot \Delta h + \frac{1}{2}\cdot \left(120\,\frac{N}{m}\right)\cdot (-0.2\,m)^{2} = 0[/tex]

The height reached by the pinball is:

[tex]\Delta h = 0.489\,m[/tex]

The distance travelled by the pinball is:

[tex]\Delta s =\frac{0.489\,m}{\sin 30^{\circ}}[/tex]

[tex]\Delta s = 0.978\,m[/tex]

To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(x,t)=Acos(kx−ωt). A transverse wave on a string is traveling in the +x direction with a wave speed of 8.50 m/s , an amplitude of 5.50×10−2 m , and a wavelength of 0.500 m . At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.52 m and t = 0.150 s .

Answers

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed  v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ   = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number =  2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2  m

and when x =  x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is  0.0549 m

A 100-g toy car moves along a curved frictionless track. At first, the car runs along a flat horizontal segment with an initial velocity of 3.33 m/s. The car then runs up the frictionless slope, gaining 0.108 m in altitude before leveling out to another horizontal segment at the higher level. What is the final velocity of the car if we neglect air resistance

Answers

Answer:

2.994 m/s

Explanation:

m = Mass of the car = 100 g

[tex]v_1[/tex] = Initial velocity = 3.33 m/s

h = Height = 0.108 m

g = Acceleration due to gravity = 9.81 m/s²

We know that energy in the system is conserved so

[tex]\dfrac{1}{2}mv_1^2=mgh+\dfrac{1}{2}mv_f^2\\\Rightarrow v_f=\sqrt{2\left({\dfrac{1}{2}v_1^2-gh}\right)}\\\Rightarrow v_f=\sqrt{2\left(\dfrac{1}{2}3.33^2-9.81\times 0.108\right)}\\\Rightarrow v_f=2.994\ m/s[/tex]

The final velocity of the car is 2.994 m/s

Final answer:

The final velocity of the car is approximately 6.05 m/s.

Explanation:

To find the final velocity of the car, we can use the principles of conservation of energy. In this case, the initial kinetic energy of the car is equal to its final potential energy. The initial kinetic energy can be calculated using the formula K.E. = 1/2 mv^2, where m is the mass of the car (100 g = 0.1 kg) and v is the initial velocity (3.33 m/s).

The final potential energy can be calculated using the formula P.E. = mgh, where m is the mass of the car, g is the acceleration due to gravity (9.8 m/s^2), and h is the change in altitude (0.108 m).

Setting the initial kinetic energy equal to the final potential energy, we can solve for the final velocity of the car:

1/2 mv^2 = mgh

Substituting the given values, we get:
1/2 (0.1 kg)(v^2) = (0.1 kg)(9.8 m/s^2)(0.108 m)

Simplifying and solving for v, we find that the final velocity of the car is approximately 6.05 m/s.

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An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in series. Initially, the switch is open and has been open for a very long time. At time t = 0 s, the switch is suddenly closed. How long after closing the switch will the potential difference across the inductor be 24 V?

Answers

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

[tex]V_L=V_oe^{-\frac{Rt}{L}}[/tex]

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

[tex]\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s[/tex]

hence, after 1.6s the inductor will have a potential difference of 24V

Final answer:

To find the time when the potential difference across the inductor will be 24 V in an LR circuit, we need to use the formula for the time-dependent current in an RL circuit, and through mathematical manipulations, substitute into the equation for the induced emf in the inductor and solve for time.

Explanation:

The time for the potential difference across the inductor to be 24V in an LR circuit can be calculated using the formula for the time-dependent current in an RL circuit, which accounts for how current evolves over time. According to Faraday's law, a changing current in an inductor generates an opposition voltage, its magnitude is determined by L (inductance of the inductor) times dI/dt (the rate of change of current).

In our case we have the equation for the current in an RL circuit when turned on: I(t) = V/R*(1 - e^(-R*t/L)) and the induced emf in the inductor V(L) = L*dI/dt. Our goal is to find the time when V(L) equals 24V. This requires substituting I(t) into the equation for V(L), setting V(L) equal to 24V, and solving for t.

This process involves the principles of RL circuits, the characteristic time constant, and Faraday's law.

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When you hear a noise, you usually know the direction from which it came even if you cannot see the source. This ability is partly because you have hearing in two ears. Imagine a noise from a source that is directly to your right. The sound reaches your right ear before it reaches your left ear. Your brain interprets this extra travel time (Δt) to your left ear and identifies the source as being directly to your right. In this simple model, the extra travel time is maximal for a source located directly to your right or left (Δt = Δtmax). A source directly behind or in front of you has equal travel time to each ear, so Δt = 0. Sources at other locations have intermediate extra travel times (0 ≤ Δt ≤ Δtmax). Assume a source is directly to your right.(a) Estimate the distance between a person's ears. (they gave us the answer of .2... apparently the program is messed up and we have to use .2(b) If the speed of sound in air at room temperature is vs = 338 m/s,find Δtmax. (Use your estimate.)
(c) Find Δtmax if instead you and the source are in seawater at the same temperature, where vs = 1534 m/s. (Use your estimate.)

Answers

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

The average distance calculated between a person's ears is 0.2 meters. The interaural time difference (ITD) is maximal when the sound source is directly to the side. For air (vs=338 m/s), Δ[tex]t_{max}[/tex] is 591.72 microseconds, and for seawater (vs= 1534 m/s), it is 130.38 microseconds. These calculations illustrate sound localization and the impact of medium on sound propagation.

Understanding Sound Localization and Interaural Time Difference

When a noise occurs directly to the right of a person, the sound waves reach the right ear before the left ear. We can calculate the maximum interaural time difference ( Δ[tex]t_{max}[/tex]) for sound reaching the ears using the given distance between the ears.

(a) The average distance between a person's ears was estimated as 0.2 meters (20 cm).

(b) To calculate Δ[tex]t_{max}[/tex] with the speed of sound in air (vs = 338 m/s), we can use the formula Δ[tex]t_{max}[/tex] = d / vs, where d is the distance between ears. Substituting the values, we get:

Δ[tex]t_{max}[/tex] = 0.2 m / 338 m/s = 0.0005917159763 seconds, or approximately 591.72 microseconds.

(c) Lastly, for sound traveling in seawater at room temperature where vs = 1534 m/s, we similarly get:

Δ[tex]t_{max}[/tex] = 0.2 m / 1534 m/s = 0.0001303763441 seconds, or approximately 130.38 microseconds.

This demonstrates the role of the medium in sound propagation and how it affects the interaural time difference.

A sinusoidal electromagnetic wave is propagating in a vacuum in the z-direction. If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 4.50 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time

Answers

Answer:

Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction

Explanation:

From maxwell's equations;

B = E/v

Where;

B is maximum magnitude of magnetic field

E is maximum electric field

v is speed of light which has a constant value of 3 x 10^(8) m/s

We are given, E = 4.5 V/m

Thus; B = 4.5/(3 x 10^(8))

B = 1.5 x 10^(-8) T

Now, for Electric field, vector E to be in the positive x-direction, the product of vector E and vector B will have to be in the positive z-direction when vector B is in the positive y-direction

Thus,

Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction

Magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex]  in the positive y-direction.

From Maxwell's equations,

[tex]\bold {B = \dfrac Ev}[/tex]

Where;

B - maximum magnitude of magnetic field = ?

E- maximum electric field  = 4.5 V/m

v- speed of light =  3 x 10^(8) m/s

Put the values in the formula,

[tex]\bold {B = \dfrac {4.5}{3 x 10^8}}\\\\\bold {B = 1.5 x 10^{-8}\ T}[/tex]

When Electric field, is in the positive x-direction, vector B is in the positive y-direction and  the product of vector E and vector B will have to be in the positive z-direction.

Therefore, magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex]  in the positive y-direction.

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Interference in 2D. Coherent red light of wavelength λ = 700 nm is incident on two very narrow slits. The light has the same phase at both slits. (a) What is the angular separation in radians between the central maximum in intensity and an adjacent maximum if the slits are 0.025 mm apart? (b) What would be the separation between maxima in intensity on a screen located 1 m from the slits? (c) What would the angular separation be if the slits were 2.5 mm apart? (d) What would the separation between maxima be on a screen 25 mm from the slits? The answer is relevant to the maximum resolution along your retina that would be useful given the size of your pupil. The spacing between cones on the retina is about 10 µm. (Of course, you would never want to look directly into a laser beam like this....) (e) How would the location of the maxima change if one slit was covered by a thin film with higher index of refraction that shifted the phase of light leaving the slit by π? Would the separation change? 3. Combined two source interference and

Answers

Answer:

Explanation:

λ

given λ = 700 nm

a) for first maxima, d*sin(θ) =  λ

sin(theta) = λ/d

= 700*10^-9/(0.025*10^-3)

= 0.028

theta θ = sin^-1(0.028)

= 1.60 degrees

b) given R = 1 m,

delta_y = λ*R/d

= 700*10^-9*1/(0.025*10^-3)

= 0.028 m or 2.8 cm

c) for first maxima, d*sin(θ) = λ

sin(θ) = λ/d

= 700*10^-9/(2.5*10^-3)

= 0.00028

theta = sin^-1(0.00028)

= 0.0160 degrees

d) R = 25 mm = 0.025 m

δ_y = λ*R/d

= 700*10^-9*0.025/(2.5*10^-3)

= 7*10^-6 m or 7 micro m

e) No. But the position of maxima and minima will be shifted.

When an earthquake occurs, two types of sound waves are generated and travel through the earth. The primary, or P, wave has a speed of about 8.0 km/s and the secondary, or S, wave has a speed of about 4.5 km/s. A seismograph, located some distance away, records the arrival of the P wave and then, 77.2 s later, records the arrival of the S wave. Assuming that the waves travel in a straight line, how far (in terms of m) is the seismograph from the earthquake?

Answers

Answer:[tex]d=7.94\times 10^5\ m[/tex]

Explanation:

Given

Speed of Primary wave [tex]v_1=8\ km/s[/tex]

Speed of secondary wave [tex]v_2=4.5\ km/s[/tex]

difference in timing of two waves are [tex]77.2\ s[/tex]

Suppose both travel a distance of d km then

[tex]t_1=\frac{d}{8}\quad \ldots (i)[/tex]

[tex]t_2=\frac{d}{4.5}\quad \ldots (ii)[/tex]

Subtract (ii) from (i)

[tex]\frac{d}{4.5}-\frac{d}{8}=77.2[/tex]

[tex]d[\frac{1}{4.5}-\frac{1}{8}]=77.2[/tex]

[tex]d[0.0972]=77.2[/tex]

[tex]d=794.23\ km[/tex]

[tex]d=7.94\times 10^5\ m[/tex]

Final answer:

To calculate the distance to the earthquake's epicenter, we use the time difference between the arrival of P-waves and S-waves and their speeds. By setting up an equation and solving for the distance, the seismograph is found to be approximately 618,940 meters from the epicenter.

Explanation:

When an earthquake occurs, two types of waves are generated: P-waves (primary waves) and S-waves (secondary waves), each with distinctive speeds. To determine the distance to the epicenter of the earthquake, we use the formula d = v × t, where d is distance, v is velocity, and t is time. Given that the P-wave has a speed of 8.0 km/s and the S-wave has a speed of 4.5 km/s, and the time difference of arrival between the two waves is 77.2 seconds, we can calculate the distance from the seismograph to the earthquake's epicenter.

Let the distance be d, then:

Time for P-wave to travel d: d / 8.0 km/sTime for S-wave to travel d: d / 4.5 km/sThe difference in travel time is 77.2 s, so: d / 4.5 km/s - d / 8.0 km/s = 77.2 s

To find the distance d, we solve the equation:

8.0×d - 4.5×d = 77.2 s × (8.0 km/s × 4.5 km/s)3.5×d = 77.2 s × 36 km²/s²d ≈ 618.94 km

Now, to convert kilometers to meters:

d ≈ 618.94 km × 1,000 m/kmd ≈ 618,940 m

Therefore, the seismograph is approximately 618,940 meters from the earthquake's epicenter.

How much heat is required, in joules, to complete the of 15.0g of water’s phase change from liquid to steam?

A. 150.7 J
B. 33,900 J
C. 62.7 J
D. 5,010 J

Answers

Answer: 62.7

Explanation:

The specific heat capacity of water is 4.186 J you multiply 4.186 and 15.0 and you will get 62.7

A space probe is sent to the vicinity of the star Capella, which is 42.2 light-years from the earth. (A light-year is the distance light travels in a year.) The probe travels with a speed of 0.9930c. An astronaut recruit on board is 19 years old when the probe leaves the earth. What is her biological age when the probe reaches Capella?

Answers

Answer:

Her current biological age = 24.02 years

Explanation:

From time dilation equation, we know that;

t = t_o * [√(1-(v²/c²))]•L_o

Where;

t = dilated time

t_o = stationary time

v = the speed of the moving object

c = the speed of light in a vacuum

First, let's convert the rest time (t_o) from light years to years.

Thus;

t_o = [c/0.993c] * [42.2]

c will cancel out and we now have;

t_o = 42.5 years

Since t = t_o * [√(1-(v²/c²))]

Thus; t = 42.5 * [√(1-(0.993²c²/c²))]

t = 42.5/[√(1 - (0.993²))]

t = 42.5 * 0.1181

t = 5.02 years

Since the astronaut was 19 years old when the probe left the earth, thus;

Her current biological age now the probe has reached Capella, will be;

19 + 5.02 = 24.02 years

A 150 g pinball rolls towards a springloaded launching rod with a velocity of 2.0 m/s
to the west. The launching rod strikes the pinball and causes it to move in the
opposite direction with a velocity of 10.0 m/s. What impulse was delivered to the
pinball by the launcher?
A 0.75 kg•m/s to the east
B 1.2 kg•m/s to the east
C 1.8 kg•m/s to the east
D 3.0 kg•m/s to the east

Answers

i believe the answer is option C 1.8 kg•m/s to the east

The impulse delivered to the pinball by the launcher is 1.8 kg•m/s to the east

Impulse:

Let us consider east as positive direction and west as negative direction.

Then, from the question, we get that the initial velocity of the pinball is

2m/s towards the west

or u = - 2 m/s

and the mass of the pinball is m = 150g = 0.15 kg

So, the initial momentum of the pinball is:

[tex]P_i=mu\\\\P_i=0.15\times(-2)\;kgm/s\\\\P_i=-0.3\;kgm/s[/tex]

Now, the final velocity of the pinball after being struck by the rod is 10  m/s towards the east,

or v = 10 m/s

So, the final momentum of the pinball is:

[tex]P_f=mu\\\\P_f=0.15\times(10)\;kgm/s\\\\P_f=1.5\;kgm/s[/tex]

Impulse is defined as the change in momentum, that is,

[tex]I=\Delta P\\\\I=P_f-P_i\\\\I=1.5-(-0.3)\\\\I=1.8\;kgm/s[/tex]

The impulse is 1.8 kgm/s towards the east.

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Which pair shows the law of reflection?

Answers

Answer:

The answer is A and C .

Explanation:

Reflection of object is reflected through the Normal (mirror) .

*incident angle = refracted angle

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