Constants Part A Two small charged spheres are 9.20 cm apart. They are moved, and the force on each of them is found to have been tripled How far apart are they now? Express your answer to three significant figures and include the appropriate units.

Answers

Answer 1

Answer:

5.3 cm

Explanation:

Let the charges on the spheres be q and Q.

r = 9.20 cm

F = k Q q / (9.20)^2 ..... (1)

Let the new distance be r' and force F'

F' = 3 F

F' = k Q q / r'^2

3 F = k Q q / r'^2 ...... (2)

Divide equation (1) by (2)

F / 3 F = r'^2 / 84.64

1 / 3 = r'^2 / 84.64

r' = 5.3 cm


Related Questions

An airplane traveling at half the speed of sound emits a sound of frequency 4.68 kHz. (a) At what frequency does a stationary listener hear the sound as the plane approaches? kHz (b) At what frequency does a stationary listener hear the sound after the plane passes? kHz

Answers

Answer:

(a) 9.36 kHz

(b) 3.12 kHz

Explanation:

(a)

V = speed of sound

[tex]v[/tex] = speed of airplane = (0.5) V

f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz

f' = Frequency heard by the stationary listener

Using Doppler's effect

[tex]f' = \frac{Vf}{V-v}[/tex]

[tex]f' = \frac{V(4680)}{V-(0.5)V)}[/tex]

f' = 9360 Hz

f' = 9.36 kHz

(b)

V = speed of sound

[tex]v[/tex] = speed of airplane = (0.5) V

f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz

f' = Frequency heard by the stationary listener

Using Doppler's effect

[tex]f' = \frac{Vf}{V+v}[/tex]

[tex]f' = \frac{V(4680)}{V+(0.5)V)}[/tex]

f' = 3120 Hz

f' = 3.12 kHz

Final answer:

The frequencies heard by the stationary listener when the airplane is approaching and when it is moving away can be calculated using the formula for the Doppler effect. The formula differs slightly depending on whether the source of the sound (in this case, the airplane) is moving towards or away from the observer (the listener).


Explanation:

This question involves the Doppler Effect, which describes how the frequency of a wave changes for an observer moving relative to the source of the wave. We are given that the airplane is moving at half the speed of sound and emits a sound of frequency 4.68 kHz.

Approaching Frequency:
When the airplane is approaching, the frequency increases. This can be calculated using the formula for the Doppler effect when the source is moving towards the observer: f' = fs * (v/(v-vs)), where v is the speed of sound, vs is the speed of the source (airplane), and fs is the source frequency. Fill in the values (remembering that the airplane's speed is half of sound speed), and solve for f' to get the approaching frequency.Receding Frequency:
When the airplane passes and is moving away, the frequency that the listener hears decreases. This can be calculated using the formula for the Doppler effect when the source is moving away from the observer: f' = fs * (v/(v+vs)). Again, fill the known values and solve for f' to find the receding frequency.
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g An ice cube has a mass of 54 g and is initially at a temperature of 0°C . The ice cube is heated until 47.4 g has become water at 100°C and 6.6 g has become steam at 100°C. How much energy (in kJ) was transferred to the ice cube to accomplish the transformation?

Answers

Answer:

55536.6 J

Explanation:

Given:

Mass of the ice = 54g

Initial temperature = 0°C

Final Temperature = 100°C

Mass of the steam = 6.6g

Now the energy required for the transformation of the ice to vapor will involve the heat requirement in the following stages as:  

1) The energy required to melt ice = mass of ice × heat of fusion of water = 54g × 334 J/g  = 18036 J

  (because heat of fusion for water = 334 J/g)

2) The energy to heat water from 0 to 100 = mass of water × specific heat of water × change in temperature = 54g × 4.186 J/g°C × 100 °C = 22604.4 J  

lastly,

3) the energy required to vaporize 6.6g of water = mass of water × heat of vaporization of water = 6.6 × 2257 J/g = 14896.2 J  

Thus,

the total energy required to transform the ice cube to accomplish the transformation = 18036 + 22604.4 + 14896.2 = 55536.6 J.

A coil is wrapped with 300 turns of wire on the perimeter of a square frame (side length = 20 cm). Each turn has the same area as the frame, and the total resistance of the coil is 1.5 Ω. A uniform magnetic field perpendicular to the plane of the coil changes in magnitude at a constant rate from 0.50 T to 0.90 T in 2.0 s. What is the magnitude of the induced emf in the coil while the field is changing?

Answers

Answer:

EMF = 2.4 V

Explanation:

As we know by Lenz law that induced EMF is given by rate of change in magnetic flux in the coil

As we know that magnetic flux is given by

[tex]\phi = NBA[/tex]

here we know that

[tex]N = 300[/tex]

[tex]Area = 0.20 \times 0.20 = 0.04 m^2[/tex]

now we for induced EMF we have

[tex]EMF = NA\frac{dB}{dt}[/tex]

here we have

[tex]EMF = (300)(0.04)(\frac{0.90 - 0.50}{2})[/tex]

[tex]EMF = 2.4 V[/tex]

Final answer:

The magnitude of the induced emf in a coil, while the magnetic field is changing, can be calculated using Faraday's law. The law states that the induced emf is equal to the negative of the rate of change of the magnetic flux linked through the circuit. We use this formula to calculate the change in magnetic field, take into accounts of coil's area and number of turns then divide by the change in time.

Explanation:

The magnitude of the induced emf in a coil, while the magnetic field is changing, can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf is equal to the negative rate of change of the magnetic fluxlinked through the circuit.

In this case, the magnetic flux, Φ, through the coil can be calculated using the formula Φ = BAN, where B is the magnetic field, A is the area of each turn (which can be calculated using the given side length) and N is the number of turns (300 in this case).

To calculate the rate of change of the magnetic flux use the formula ΔΦ/Δt, where ΔΦ is the change in magnetic flux and Δt is the change in time. In this case, the magnetic field changes from 0.50 T to 0.90 T in 2.0 s, so the change in magnetic flux is (B_final - B_initial)*A*N, and Δt is 2.0 s.

The magnitude of the induced emf (which is also the negative rate of change of the magnetic flux) is given by Faraday's law, |ε| = |ΔΦ/Δt|.?

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For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser pointer (λ = 685 nm) through the slit onto a screen 5.4 m away, a diffraction pattern appears. The bright band in the center of the pattern is 7.0 cm wide. What is the width of the slit?

Answers

Answer:

105.6857 nm

Explanation:

Wave length of laser pointer = λ = 685 nm = 685×10⁻⁹ m

Distance between screen and slit = L = 5.4 m

Width of bright band = 2x = 7 cm

⇒x = 3.5 cm = 3.5×10⁻² m

The first minimum occurs at

sin θ = λ/d

θ = λ/d                 (since the angle is very small, sin θ ≈ θ)

Width of slit

d = λL/x

⇒d = 685×10⁻⁹×5.4/3.5×10⁻²

⇒d = 1056.857×10⁻⁷ m

∴ The width of the slit is 105.6857 nm

A baseball bat is 32 inches (81.3 cm) long and has a mass of 0.96 kg. Its center of mass is 22 inches (55.9 cm) from the handle end. You hold the bat at the very tip of the handle end (the knob) and let it swing in simple harmonic motion. What is the bat’s moment of inertia if its period of oscillation is 1.35 seconds?

Answers

Answer:

0.24 kgm²

Explanation:

[tex]L[/tex] = length of the bat = 81.3 cm = 0.813 m

[tex]m[/tex]  = mass of the bat = 0.96 kg

[tex]d[/tex]  = distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m

[tex]T[/tex]  = Period of oscillation = 1.35 sec

[tex]I[/tex] = moment of inertia of the bat

Period of oscillation is given as

[tex]T = 2\pi \sqrt{\frac{I}{mgd}}[/tex]

[tex]1.35 = 2(3.14) \sqrt{\frac{I}{(0.96)(9.8)(0.559)}}[/tex]

[tex]I[/tex] = 0.24 kgm²

Suppose I launch a charged particle into a uniform magnetic field. The particle will trace out a circular path in the field. Which of the following changes will decrease the radius of the circular path? Decrease the strength of the magnetic field Increase the mass of the particle Increase the charge on the particle Increase the volume of the particle Increase the speed of the particle

Answers

Answer:

increase the charge

Explanation:

The force acting on charge particle when moving perpendicular to the magnetic field

F = q v B

The centripetal force is given by

F =  mv^2 / r

Comparing both, we get

r = m v / B q

That means the radius of the circular path depends on mass of the charge particle, velocity of the charge particle, magnetic field strength and charge of the particle.

To decrease the radius:

1. increase the charge

2. increase the magnetic field strength

3. decrease the speed

4. decrease the mass

The factor that will decrease the radius of the circular path is increase the charge on the particle.

Magnetic force on the charged particle

The magnetic force on the charged particle is calculated as follows;

F = qvB

Where;

q is the chargev is the velocity of the chargeB is magnetic field

Centripetal force on the charged particle

The force on the particle moving in circular path is given as;

F = mv²/r

qvB = mv²/r

qB = mv/r

r = mv/qB

Thus, the factor that will decrease the radius of the circular path is increase the charge on the particle.

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An ideal gas experiences a change of internal energy of 284 J/kg when its temperature is increased by 12oC (at a constant volume). What is its specific heat

Answers

Answer:

23.5 J/Kg °C

Explanation:

The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree Celcius.

c = Q / m ΔT

Here, Q / m = 282 J/kg, ΔT = 12

So, the specific heat = 282 / 12 = 23.5 J/Kg °C

A Boeing 777 aircraft has a mass of 300,000 kg. At a certain instant during its landing, its speed is 27.0 m/s. If the braking force is 445,000 N, what is the speed of the airplane 10.0 s later?

Answers

Answer:

Speed of the airplane 10.0 s later = 12.2 m/s

Explanation:

Mass of Boeing 777 aircraft = 300,000 kg

Braking force = 445,000 N

Deceleration

            [tex]a=\frac{445000}{300000}=1.48m/s^2[/tex]

Initial velocity, u = 27 m/s

Time , t = 10 s

We have equation of motion, v =u +at

            v = 27 + (-1.48) x 10 = 27 - 14.8 = 12.2 m/s

Speed of the airplane 10.0 s later = 12.2 m/s

Final answer:

After applying the deceleration due to the braking force for 10 seconds, the speed of the Boeing 777 airplane reduces to 12.17 m/s.

Explanation:

Calculating the Final Speed of a Boeing 777 After Braking

To determine the speed of a Boeing 777 aircraft after a 10-second interval of braking, we will use the relationship between force, mass, initial velocity, and acceleration provided by Newton’s second law of motion. The braking force applied on the aircraft is 445,000 N acting opposite to the direction of motion. We can calculate the deceleration using the formula:

F = ma

Where F is the force, m is the mass, and a is the acceleration (deceleration in this case since the force is applied opposite to the direction of motion). The mass (m) of the airplane is 300,000 kg, so we can solve for 'a' as follows:

a = F / m = 445,000 N / 300,000 kg = 1.483 m/s² (deceleration)

Next, we use the kinematic equation to find the final velocity (vf) after 10 seconds:

vf = vi + at

Where vi is the initial velocity, and t is the time. We know that the initial velocity (vi) is 27.0 m/s and the time (t) is 10.0 s.

Plugging the numbers in:

vf = 27.0 m/s - (1.483 m/s² × 10.0 s) = 27.0 m/s - 14.83 m/s = 12.17 m/s

Therefore, the speed of the Boeing 777 airplane 10 seconds later is 12.17 m/s.

(a) A uniform disk of mass 21 kg, thickness 0.5 m, and radius 0.6 m is located at the origin, oriented with its axis along the axis. It rotates clockwise around its axis when viewed from above (that is, you stand at a point on the axis and look toward the origin at the disk). The disk makes one complete rotation every 0.7 s. What is the rotational angular momentum of the disk? What is the rotational kinetic energy of the disk? (Express your answer for rotational angular momentum in vector form.)

Answers

Answer:L=33.93

R.E.=152.30

Explanation:

Given

mass of disk[tex]\left ( m\right )[/tex]=21kg

thickness[tex]\left ( t\right )[/tex]=0.5m

radius[tex]\left ( r\right )[/tex]=0.6m

t=0.7 sec for every complete rotation

therefore angular velocity[tex]\left ( \omega \right ) =\frac{\Delta \theta }{\Delta t}[/tex]

[tex]\left ( \omega \right )=\frac{ 2\pi }{0.7}=8.977[/tex] rad/s

Rotational angular momentum is given by

[tex]L=I\omega [/tex]

[tex]I[/tex]=[tex]\frac{mr^2}{2}[/tex]

I=3.78 [tex]kg-m^2[/tex]

[tex]L=3.78\times 8.977[/tex]

L=33.93 [tex]\hat{j}[/tex]

considering disk is rotating in z-x plane

Rotational kinetic energy=[tex]\frac{I\omega ^{2}}{2}[/tex]=152.30 J

Final answer:

The rotational angular momentum of the disk is 33.89 kg.m²/s directed into the page, and the rotational kinetic energy of the disk is 151.5 J.

Explanation:

To calculate the rotational angular momentum of the disk, we will first need to compute the moment of inertia and angular velocity. The moment of inertia is given by the formula for a solid disk, which is I = 1/2mR². Here, m is the mass of the disk (in this case, 21 kg), and R is the radius of the disk (which is 0.6 m). So, I = 0.5*21*(0.6)² = 3.78 kg.m².

Furthermore, the angular velocity (ω) can be calculated using the formula 2π/T, where T is the period, which is the time it takes for one complete rotation (0.7 s in this situation). Therefore, ω = 2π/0.7 = 8.96 rad/s.

Finally, the rotational angular momentum (L), is given by the product of the moment of inertia and angular velocity, L = I*ω. Therefore, L = 3.78 kg.m² * 8.96 rad/s = 33.89 kg.m²/s.

The direction of the rotational angular momentum is along the axis of rotation, which depends on the right-hand rule. Since it is rotating clockwise when viewed from above, the vector is directed into the page (negative z-direction).

To compute the rotational kinetic energy (KE), we use the formula KE = 1/2 I ω². Substituting the values, we get KE = 0.5*3.78 kg.m²*(8.96 rad/s)² = 151.5 J.

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During a volleyball serve, the ball leaves the hand with an initial velocity of 10 m/s angled 41 degrees from the horizontal. What are the horizontal and vertical velocities of the ball?

Answers

Answer:

7.55 m/s, 6.56 m/s

Explanation:

v = 10 m/s, theta = 41 degree

Horizontal component of velocity = v x Cos theta = 10 x Cos 41 = 7.55 m/s

Vertical component of velocity = v x Sin theta = 10 x Sin 41 = 6.56 m/s

Final answer:

The ball's horizontal velocity is approximately 7.57 m/s, and its vertical velocity is approximately 6.46 m/s immediately after the serve.

Explanation:

The student's question about the initial horizontal and vertical velocities of a volleyball serve involves breaking down the initial velocity into its component parts using trigonometric functions. Given an initial velocity (v) of 10 m/s and an angle (θ) of 41 degrees from the horizontal, the horizontal component (vx) is calculated using cosine, and the vertical component (vy) is calculated using sine.

Using the formula:

vx = v ∙ cos(θ)

vy = v ∙ sin(θ)

For this problem:

vx = 10 m/s ∙ cos(41°)

vy = 10 m/s ∙ sin(41°)

Plugging in the values yields:

vx ≈ 7.57 m/s

vy ≈ 6.46 m/s

The ball's horizontal velocity is approximately 7.57 m/s, and its vertical velocity is approximately 6.46 m/s immediately after the serve.

A wind turbine has 12,000 kg blades that are 37 m long. The blades spin at 24 rpm .If we model a blade as a point mass at the midpoint of the blade, what is the inward force necessary to provide each blade's centripetal acceleration

Answers

Answer:

[tex]F_c = 1.4 \times 10^6 N[/tex]

Explanation:

As we know

[tex]f = 24 rpm[/tex]

so we will have

[tex]f = 24 \frac{1}{60} hz = 0.4hz[/tex]

now angular frequency is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 0.8\pi[/tex]

Now the inwards is given as centripetal force

[tex]F_c = m\omega^2 r[/tex]

[tex]F_c = (12000)(0.8\pi)^2(\frac{37}{2})[/tex]

[tex]F_c = 1.4 \times 10^6 N[/tex]

Final answer:

The inward force necessary to provide each blade's centripetal acceleration is approximately 4,755,076 Newtons.

Explanation:

To calculate the inward force necessary to provide each blade's centripetal acceleration, we can use the formula for centripetal force:

[tex]Fc = m \times \omega^2 \times r[/tex]

where Fc is the centripetal force, m is the mass of the blade, ω is the angular velocity, and r is the radius of the blade.

In this case, the mass of the blade is 12,000 kg, the angular velocity is 24 rpm (which can be converted to radians per second by multiplying by 2π/60), and the radius of the blade is 37 m.

Plugging these values into the formula, we get:

Fc = 12000 kg * (24 * 2π/60)^2 * 37 m

Fc ≈ 4,755,076 N

So, the inward force necessary to provide each blade's centripetal acceleration is approximately 4,755,076 Newtons.

The left side of the heart creates a pressure of 120 mm Hg by exerting a force directly on the blood over an effective area of 15.0 cm2. What force does it exert to accomplish this

Answers

Answer:

23.99N

Explanation:

Given:

Pressure created by the heart = 120mm of hg

converting the pressure into the standard unit of N/m²

1mm of hg = [tex]\frac{1}{760}atm=\frac{1}{760}\times 1.013\times10^5N/m^2[/tex]

now, 120 mm of hg in N/m² will be

120mm of hg = [tex]\frac{12}{760}\times 1.013\times10^5N/m^2[/tex]

also

given effective area = 15.0 cm² = 15 × 10⁻⁴m²

Now,

Force = Pressure × Area

thus,

Force exerted will be =  [tex]\frac{120}{760}\times 1.013\times10^5N/m^2[/tex] ×  15 × 10⁻⁴m²

or

Force exerted will be = 23.99N

The force that it exerts is about 24.0 N

[tex]\texttt{ }[/tex]

Further explanation

Let's recall Hydrostatic Pressure formula as follows:

[tex]\boxed{ P = \rho g h}[/tex]

where:

P = hydrosatic pressure ( Pa )

ρ = density of  fluid ( kg/m³ )

g = gravitational acceleration ( m/s² )

h = height of a column of liquid ( m )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

blood pressure = P = 120 mmHg = 0.12 mHg

effective area = A = 15.0 cm² = 15.0 × 10⁻⁴ m²

density of mercury = ρ = 13600 kg/m³

gravitational acceleration = g = 9.8 m/s²

Asked:

force = ?

Solution:

We will use this following formula to solve this problem:

[tex]P = F \div A[/tex]

[tex]F = P A[/tex]

[tex]F = \rho g h A[/tex]

[tex]F = 13600 \times 9.8 \times 0.12 \times ( 15.0 \times 10^{-4} )[/tex]

[tex]F \approx 24.0 \texttt{ N}[/tex]

[tex]\texttt{ }[/tex]

Conclusion :

The force that it exerts is about 24.0 N

[tex]\texttt{ }[/tex]

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[tex]\texttt{ }[/tex]

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter when there is no flow is 3 m (gauge). Find the rate of flow for which the throat pressure will be 2m of water absolute. Discharge coefficient for the meter is 0.97.

Answers

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

[tex]Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}[/tex]

Where,

[tex]C_d[/tex] is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = [tex]\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2[/tex]

A₂ = Area at the throat

A₂ = [tex]\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2[/tex]

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

[tex]Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}[/tex]

or

[tex]Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}[/tex]

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

The North American Plate is moving west at a rate of approximately 20 mm/yr. How long will it take for New York to move 10° longitude west of its current position? (Assume 110 km/°longitude at New York's latitude.)

Answers

Answer:

55000000 years

Explanation:

Rate of North American Plate moving (Velocity) = 20 mm/yr

Degrees New York has to move = 10° west

New York's longitude = 110 km/°

Distance of New York = 110×10 = 1100 km

= 1100×10⁶ mm

[tex]\text {Time taken by the continental plate}=\frac {\text {Distance of New York}}{\text {Rate of North American Plate moving (Velocity)}}\\\Rightarrow \text {Time taken by the continental plate}=\frac{1100\times 10^6}{20}\\\Rightarrow \text {Time taken by the continental plate}=55\times 10^6\ years[/tex]

∴ Time taken by New York to move 10° west is 55000000 years

A flat sheet of ice has a thickness of 2.20 cm. It is on top of a flat sheet of crystalline quartz that has a thickness of 1.50 cm. Light strikes the ice perpendicularly and travels through it, and then through the quartz. In the time it takes the light to travel through the two sheets, how far would it have traveled in a vacuum?

Answers

Answer:

[tex]Distance_{vaccum}=5.19cm[/tex]

Explanation:

The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under

Total time = Time taken through ice + Time taken through quartz

Time taken through ice = Thickness of ice / (speed of light in ice)

[tex]T_{ice}=\frac{2.20\times 10^{-2} \times \mu _{ice}}{V_{vaccum}}[/tex]

[tex]T_{quartz}=\frac{1.50\times 10^{-2} \times \mu _{quartz}}{V_{vaccum}}[/tex]

Thus in the same time the it would had covered a distance of

[tex]Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}][/tex]

we have

[tex]\mu _{ice}=1.309\\\\\mu _{quartz}=1.542[/tex]

Applying values we have

[tex]Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542][/tex]

[tex]Distance_{vaccum}=5.19cm[/tex]

Final answer:

The light, after going through ice and quartz, would have traveled approximately 5.198 cm in vacuum. This is calculated considering indices of refraction for each material and their thicknesses.

Explanation:

The subject of this question is in the domain of Physics, specifically optics. The problem wants us to compare the distances light travels in different media to its travel in vacuum. The key to solve this matter is knowing the speed of light changes depending on the medium it passes through: it's slower in media like ice or quartz than in vacuum, due to these materials' indices of refraction.

The index of refraction for ice is approximately 1.31 and for crystalline quartz is approximately 1.544. The index of refraction (n) is the ratio of the speed of light in vacuum (c), to the speed of light in the medium (v), or n = c/v. To find out the distances light travels in vacuum for the time it takes to travel through the ice and quartz, we need to multiply the thickness of each material by their respective indices of refraction.

For the ice, it's 2.20 cm * 1.31 = 2.882 cm and for the quartz, it's 1.50 cm * 1.544 = 2.316 cm. So the total distance light would have traveled in vacuum in the same time it passes through both sheets is approximately 2.882 cm + 2.316 cm = 5.198 cm.

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A rod (length = 80 cm) with a rectangular cross section (1.5 mm × 2.0 mm) has a resistance of 0.20 Ω. What is the resistivity of the material used to make the rod?

Answers

Answer:

The resistivity of the material used to make the rod is ρ= 7.5 * 10⁻⁷ Ω.m

Explanation:

R= 0.2 Ω

L= 0.8 m

S= 1.5mm*2mm= 3 mm² = 3 * 10⁻⁶ m²

ρ = (R*S)/L

ρ= 7.5 * 10⁻⁷ Ω.m

The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic friction between the block and the plane is μk = 0.24. Determine the distance the block will slide before it stops.

Answers

Answer:

The distance the block will slide before it stops is x= 3.4 m .

Explanation:

m1= 0,02 kg

V1= 400 m/s

m2= 2 kg

V2= ?

g= 9.8 m/s²

μ= 0.24

N = m2 * g = 19.6 N

Fr= μ * N

Fr= 4.704 N

due to the conservation of the amount of movement:

m1*V1 = m2*V2

V2= 4 m/s

Fr = m2*a

a= Fr/m2

a= -2.352 m/s²

matching momentum and amount of movement

Fr*t=m2*V2

t= 1.7 sec

x= V2*t - (a*t²)/2

x= 3.4 m

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

[tex]v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f[/tex]

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

[tex]S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2[/tex]

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

A rope with a mass density of 1 kg/m has one end tied to a vertical support. You hold the other end so that the rope is horizontal and has a tension of 4 N. If you move the end of the rope back and forth, you produce a transverse wave in the rope with a wave speed of 2 m/s. If you double the amount of tension you exert on the rope, what is the wave speed?

Answers

Answer:

[tex]v' = 2.83 m/s[/tex]

Explanation:

Velocity of wave in stretched string is given by the formula

[tex]v = \sqrt{\frac{T}{\mu}}[/tex]

here we know that

T = 4 N

also we know that linear mass density is given as

[tex]\mu = 1 kg/m[/tex]

so we have

[tex]v = \sqrt{\frac{4}{1}} = 2 m/s[/tex]

now the tension in the string is double

so the velocity is given as

[tex]v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s[/tex]

[tex]v' = 2.83 m/s[/tex]

Answer True or False to each case. Coulomb's law states that the force between static charges Q1 and Q2 is K Q1Q2/r2 where K is a universal constant. is a force of attraction. may be attractive or repulsive. is K Q1Q2/r where K is a universal constant. is proportional to 1/r2.

Answers

Explanation:

The Coulomb's law states that the force acting on two charges is directly proportional to the product of charges and inversely proportional to the square of distance between them . Mathematically, it is given by

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Where

k is the electrostatic constant

q₁ and q₂ are charges

r is the distance between them.

The SI unit of electric force is Newton. It can be attractive or repulsive. The attraction or repulsion depend on charges. If both charges are positive, the force is repulsive and if both are opposite charges, the force is attractive.

what is the internal energy of 50 moles of neon gas at 20.0 degrees celsius

Answers

Answer:

[tex]U = 1.83 \times 10^5 J[/tex]

Explanation:

Total internal energy of the gas is given by the formula

[tex]U = \frac{f}{2}nRT[/tex]

here as we know that neon gas is monoatomic gas

so we will have

[tex]f = 3[/tex]

n = 50 moles

[tex]T = 20 ^0C = 20 + 273 = 293 K[/tex]

now from above equation we will have

[tex]U = \frac{3}{2}(50)(8.31)(293)[/tex]

[tex]U = 1.83 \times 10^5 J[/tex]

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?

Answers

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

For an electron moving in a direction opposite to the electric field A) both its potential energy and itelectric potential remain constant its potential energy decreases and itselectricpotential increases. its potential energy decreases and its electric potential decreases. its potential energy increases and itselectric potential decreases. its potential energy increases and its electric potential increases

Answers

Answer:

Its potential energy decreases and itselectricpotential increases.

Explanation:

The electric potential does not depend on the charge but only on the magnitude of the electric field. In particular:

- Electric potential decreases when moving in the same direction of the electric field lines

- Electric potential increases when moving in the opposite direction to the field lines

So in this case, since the electron is moving in a direction opposite to the field, the electric potential increases.

However, the electric potential energy of a charge is given by

[tex]U=qV[/tex]

where

q is the charge

V is the electric potential

Here we said that the electric potential is increasing: however, the charge q of an electron is negative. This means that the product (qV) is increasing in magnitude but it is negative, so the potential energy of the electron is decreasing.

For an electron moving in a direction opposite to the electric field its potential energy decreases and its electric potential increases.

What is electric potential?

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field.it is denoted by U.

However, the electric potential energy of a charge is given by

[tex]\rm{U=qv}[/tex]

q is denoted for the charge.

V is denoted for the electric potential.

The above relation shows that the electric potential is inversely proportional to the electric potential energy.

The electric potential is independent of the charge. but it depends on the direction in which the charge is moving. If the charge moves in the direction of the electric field electric potential decreases and vice versa.

Here the electron is moving in a direction opposite to the field, the electric potential increases while electric potential energy decreases.

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A​ right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 ​lb/ft3. How much work is required to pump all of the fuel to a point 13 ft above the top of the​ tank?

Answers

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ [tex]ft^{3}[/tex] to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = [tex]\pi r^{2}[/tex] =  [tex]\pi 4^{2}[/tex] =16[tex]\pi[/tex] [tex]ft^{2}[/tex]

Now,

Work done required = [tex]\int_{0}^{8} 52\times 16\pi (21 - y)dy[/tex]

                                  = [tex]832\pi \int_{0}^{8} (21 - y)dy[/tex]

                                  = 832[tex]\pi([/tex][tex][ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\[/tex])

                                  = 113152[tex]\pi[/tex] = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 2.15 x 10^5 kg train moving at 0.710 m/s. (a) What is the force constant of the spring? (N/m)
(b) What speed would the train be going if it only compressed the spring 0.640 m?
(c) What force does the spring exert when compressed 0.640 m? (Enter the magnitude only.)

Answers

Answer:

(a) the constant of the spring is 27095.375 J/m2

(b) 0.052 m/s

(c) F=17341.04 N

Explanation:

Hello

The law of conservation of energy states that the total amount of energy in any isolated system (without interaction with any other system) remains unchanged over time.

if we assume that there is no friction, then, the kinetic energy of the train (due to movement) will be equal to the energy accumulated in the spring

Step 1

energy of the train (kinetic)

[tex]E_{k}=\frac{m*v^{2} }{2}\\  E_{k}=\frac{2.15*10^{5}kg*(0.71\frac{m}{s}) ^{2} }{2}\\E_{k}=54190.75 J\\ \\[/tex]

step 2

energy of the spring

[tex]E_{s}=\frac{Kx^{2} }{2}\\[/tex]

where K is the constant of the spring and x the length compressed.

[tex]E_{s}=\frac{Kx^{2} }{2}=54190.75 J\\\frac{Kx^{2} }{2}=54190.75 J\\\\k=\frac{2*54190.75 J}{x^{2}}\\k=\frac{2*54190.75 J}{(2.0 m)^{2} }\\ k=27095.375\frac{J}{m^{2} } \\\\[/tex]

(a) the constant of the spring is 27095.375 J/m2

(b)

[tex]x=0.640 m\\\\E_{s}=\frac{27095.375 (\frac{j}{m^{2} })*(0.640m)^{2}  }{2}\\ E_{s}=5549.1328 J\\[/tex]

equal to train energy

[tex]E_{k}=\frac{2.15*10^{5}kg*(v) ^{2} }{2}\\\frac{2.15*10^{5}kg*(v) ^{2} }{2}=5549.1328 J\\v^{2}=\frac{2*5549.1328 J}{2.15*10^{5}kg}\\v=0.052 \frac{m}{s} \\\\[/tex]

(b) 0.052 m/s

(c)

[tex]F= kx\\\\F=27095.375 \frac{J}{m^{2} }*(0.640 m)\\ F=17341.04 N\\F=17341.04[/tex]

(c) F=17341.04 N

I hope it helps

Final answer:

The force constant of the spring is 38.03 N/m. If the train compressed the spring 0.640 m, the train would be going at a speed of 0.418 m/s. The force the spring exerts when compressed 0.640 m is 24.34 N.

Explanation:

To solve these questions we use conservation of energy principles and the spring equation, F = k*x. The initial kinetic energy of the train is converted into potential energy in the spring at the point of maximum compression.

(a) What is the force constant of the spring?
Here we equate kinetic energy with potential energy by using the equations KE = 0.5 * m * v^2 and PE = 0.5 * k * x^2. We can solve for k to get k = m * v^2 / x^2, so it's (2.15 x 10^5 kg * (0.710 m/s)^2)/(2.00 m)^2 = 38.03 N/m.

(b) What speed would the train be going if it only compressed the spring 0.640 m?
Here we rearrange the previous equation for velocity, v = sqrt(k * x^2 / m) which we then place values into to calculate v = sqrt(38.03 N/m * (0.640 m)^2)/(2.15 x 10^5 kg) = 0.418 m/s.

(c) What force does the spring exert when compressed 0.640 m?
Here we refer back to Hooke's Law (F = k*x), which states that the force required to compress or extend a spring by some distance x is proportional to that distance. The spring constant (k) is the proportionality constant in this relationship. So, the force is F = (38.03 N/m * 0.640 m) = 24.34 N.

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Two objects attract each other gravitationally. If the distance between their centers increases by a factor of 3, and both of the objects' masses increase by the same factor, how does the gravitational force between them change? A) The gravitational force decreases by a factor of 3.
B) The gravitational force increases by a factor of 3.
C) The gravitational force decreases by a factor of 9.
D) The gravitational force increases by a factor of 9.
E) The gravitational force remains unchanged.

Answers

Answer:

E

Explanation:

F = G * m1 * m2 / r^2     Increase the distance by 3

F1 = G * m1 * m2 / (3r)^2

F1 = G * m1 * m2 / (9*r^2) What this means is the the force decreases by a factor of 9, but we are not done.

F2 = G * 3m1 * 3m2 / (9 r^2)

F2 = G * 9 m1 * m2 / (9 r^2)

In F2 the 9s cancel out and we are left with

F2 = G * m1 * m2/r^2 which is the same thing.

F2 equals F

Answer:

E. The gravitational force remains unchanged

Explanation:

A wire 73.6 cm long carries a 0.720 A current in the positive direction of an x axis through a magnetic field with an x component of zero, a y component of 0.000270 T, and a z component of 0.00770 T. Find the (a) x, (b) y, and (c) z components of the magnetic force on the wire.

Answers

The magnetic force on the wire has an x-component of 0 N, a y-component of 0.004056224 N in the negative y-direction, and a z-component of 0.000143712 N in the positive z-direction.

The question involves calculating the magnetic force on a current-carrying wire in a magnetic field using the right-hand rule and the formula F = I × (L × B), where F is the force, I is the current, L is the length of the wire, and B is the magnetic field. We know the current I is 0.720 A, the length L is 73.6 cm (which we will convert to meters), and the magnetic field components are By = 0.000270 T and Bz = 0.00770 T.

First, let's convert the length of the wire from centimeters to meters: L = 73.6 cm = 0.736 m.

The force on the wire in the x, y, and z directions (Fx, Fy, Fz) can be calculated using the cross product of the current direction (along the x-axis) and the magnetic field components. Since there is no x component for the magnetic field (Bx = 0), the force in the x-direction (Fx) will be zero.

Using the right-hand rule, the force in the y-direction (Fy) will be:

Fy = I × (L × Bz) = 0.720 A × (0.736 m × 0.00770 T) = 0.004056224 N, pointing in the negative y-direction (since the current is in the positive x-direction and Bz is positive).

Similarly, the force in the z-direction (Fz) is calculated as:

Fz = I × (L × By) = 0.720 A × (0.736 m × 0.000270 T) = 0.000143712 N, pointing in the positive z-direction.

A clarinetist, setting out for a performance, grabs his 3.230 kg3.230 kg clarinet case (including the clarinet) from the top of the piano and carries it through the air with an upward force of 30.10 N.30.10 N. Find the case's vertical acceleration. Indicate an upward acceleration as positive and a downward one as negative.

Answers

Answer:

- 0.5 m/s²

Explanation:

m = mass of the clarinet case = 3.230 kg

W = weight of the clarinet case in downward direction

a = vertical acceleration of the case

Weight of the clarinet case is given as

W = mg

W = 3.230 x 9.8

W = 31.654 N

F = Upward force applied = 30.10 N

Force equation for the motion of the case is given as

F - W = ma

30.10 - 31.654 = 3.230 a

a = - 0.5 m/s²

Final answer:

By applying Newton's second law and accounting for the forces acting on the clarinet case, it is found to experience a downward vertical acceleration of 0.489 m/s², due to the net force acting on it being in the downward direction.

Explanation:

To find the vertical acceleration of the clarinet case, first identify the forces acting on it. The force of gravity (weight) pulls it downward, which can be calculated by multiplying the mass of the case by the acceleration due to gravity (9.81 m/s²). The equation for weight is W = mg, where m is mass and g is gravity.

Substituting the given values, W = 3.230 kg * 9.81 m/s² = 31.68 N. The net force [tex]F_{net}[/tex] acting on the case is the upward force by the clarinetist subtracting the weight of the case, [tex]F_{net}[/tex] = 30.10 N - 31.68 N = -1.58 N. Applying Newton's second law, F = ma, and solving for acceleration (a), a = [tex]F_{Net/m}[/tex], we find the case's vertical acceleration as a = -1.58 N / 3.230 kg = -0.489 m/s².

Therefore, the case experiences a downward acceleration of 0.489 m/s², indicating that it is slowing in its ascent or accelerating downward.

A tennis racket hits a tennis ball with a force of F=at−bt2, where a = 1290 N/ms , b = 330 N/ms2 , and t is the time (in milliseconds). The ball is in contact with the racket for 2.80 ms . If the tennis ball has a mass of 55.6 g , what is the resulting velocity of the ball, v, after the ball is hit by the racket?

Answers

The resulting velocity of the tennis ball is calculated by first determining the impulse, which is the integral of the force over the contact time. The impulse is then equal to the change in momentum, allowing us to solve for the velocity. The calculated velocity of the ball after impact is 74.67 m/s.

To calculate the resulting velocity of the ball after being hit by the racket, we first need to determine the impulse imparted to the ball. The force applied by the racket is variable and given by F=at-bt², where a = 1290 N/ms, b = 330 N/ms², and t is the time in milliseconds. To calculate the impulse (J), we integrate the force over the contact time, from 0 to 2.80 ms.

Impulse, J, is the integral of F with respect to t, which gives us J = (1/2)at² - (1/3)bt³ evaluated from 0 to 2.80 ms. Plugging in the values:

J = (1/2)(1290 N/ms)(2.80 ms)² - (1/3)(330 N/ms²)(2.80 ms)³ = 1290(3.92) - 330(2.744) N = 5056.8 - 905.52 N = 4151.28 Nms

The impulse is equal to the change in momentum of the tennis ball. Considering the ball's mass m = 55.6 g = 0.0556 kg, and initial velocity, u = 0 m/s (since it's hit from rest), we apply the impulse-momentum theorem:

J = F - mv, therefore, v = J/m.

Substituting the values we have: v = 4151.28 Nms / 0.0556 kg = 74667.7 m/s = 74.67 m/s.

So, the resulting velocity of the tennis ball is 74.67 m/s after the racket's impact.

The resulting velocity of the ball after being hit by the racket is approximately [tex]\( 49466.325 \, \text{m/s} \)[/tex].

To find the resulting velocity of the ball after being hit by the racket, we can use the impulse-momentum theorem, which states that the change in momentum of an object is equal to the impulse applied to it:

[tex]\[ J = \Delta p \][/tex]

The impulse J is equal to the integral of the force F with respect to time t over the duration of the contact:

[tex]\[ J = \int_{0}^{2.80} F \, dt \][/tex]

Given that [tex]\( F = at - bt^2 \)[/tex], we can integrate [tex]\( F \)[/tex] with respect to [tex]\( t \)[/tex] to find [tex]\( J \)[/tex]:

[tex]\[ J = \int_{0}^{2.80} (at - bt^2) \, dt \][/tex]

[tex]\[ J = \left[ \frac{1}{2} at^2 - \frac{1}{3} bt^3 \right]_{0}^{2.80} \][/tex]

[tex]\[ J = \left( \frac{1}{2} a(2.80)^2 - \frac{1}{3} b(2.80)^3 \right) - \left( \frac{1}{2} a(0)^2 - \frac{1}{3} b(0)^3 \right) \][/tex]

[tex]\[ J = \left( \frac{1}{2} \cdot 1290 \cdot (2.80)^2 - \frac{1}{3} \cdot 330 \cdot (2.80)^3 \right) - 0 \][/tex]

[tex]\[ J = \left( \frac{1}{2} \cdot 1290 \cdot 7.84 - \frac{1}{3} \cdot 330 \cdot 21.952 \right) \][/tex]

[tex]\[ J = \left( 5056.64 - 2308.656 \right) \][/tex]

[tex]\[ J = 2747.984 \, \text{N} \cdot \text{ms} \][/tex]

Now, we know that impulse [tex]\( J \)[/tex] is also equal to the change in momentum [tex]\( \Delta p \)[/tex]  of the ball:

[tex]\[ J = \Delta p = mv - mu \][/tex]

Where:

- [tex]\( m \)[/tex] is the mass of the ball,

- v is the final velocity of the ball,

- u is the initial velocity of the ball (which we assume to be zero as the ball is initially at rest).

So, we can rearrange the equation to solve for v:

[tex]\[ v = \frac{J}{m} \][/tex]

Now, let's plug in the values to find v:

[tex]\[ v = \frac{2747.984 \, \text{N} \cdot \text{ms}}{0.0556 \, \text{kg}} \][/tex]

[tex]\[ v \approx 49466.325 \, \text{m/s} \][/tex]

So, the resulting velocity of the ball after being hit by the racket is approximately [tex]\( 49466.325 \, \text{m/s} \)[/tex].

An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electron is 4.52×105 m/s pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.395 m east of point A?

Answers

Final answer:

The speed of the electron when it reaches point B, which is 0.395 m east of point A, is approximately 3.86×10⁵ m/s, directed towards the east.

Explanation:

The initial speed of the electron is provided as 4.52×10⁵ m/s. Given the electric field, we can calculate the force on the electron as F = qE, where q is the charge of the electron (-1.60 × 10⁻¹⁹ C) and E is the electric field (1.55 N/C). Hence, the force acting on the electron is F = -1.60 × 10⁻¹⁹ C * 1.55 N/C = -2.48 × 10⁻¹⁹ N.

Using F = ma, we can calculate the acceleration of the electron. Knowing the mass of the electron is 9.11 × 10⁻³¹ kg, the acceleration is a = F/m = -2.48 × 10⁻¹⁹ N / 9.11 × 10⁻³¹ kg = -2.72 × 10¹¹ m/s².

Given the distance of the movement is 0.395 m, we can use the equation v² = u² + 2as to solve for the final velocity 'v', where 'u' is the initial velocity, 'a' is the acceleration and 's' is the distance. Substituting the known values, we get v = sqrt((4.52×10⁵ m/s)² - 2 * 2.72 × 10¹¹ m/s² * 0.395 m) ≈ 3.86 x 10⁵ m/s (approximately).

So, the speed of the electron when it reaches point B is approximately 3.86 x 10⁵ m/s, directed towards the east.

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A person is standing a distance D = 5.8 m in front of a flat, vertical mirror. The distance from the ground to his eyes is H = 1.6 m. An object is placed on the ground a distance d = D/2 = 2.9 m in front of the mirror. At what height h should the bottom of the mirror be so that the person can see the bottom of the object?

Answers

Answer:

it is at height of y = 0.533 m from ground

Explanation:

As per law of reflection we know that angle of incidence = angle of reflection

so here we have

[tex]tan\theta_i = tan\theta_r[/tex]

here we know that

[tex]tan\theta_i = \frac{y}{d}[/tex]

also we have

[tex]tan\theta_r = \frac{H - y}{D}[/tex]

now we have

[tex]\frac{H - y}{D} = \frac{y}{d}[/tex]

here we have

[tex]\frac{1.6 - y}{5.8} = \frac{y}{2.9}[/tex]

[tex]3y = 1.6[/tex]

[tex]y = 0.533 m[/tex]

Final answer:

To see the bottom of the object in the mirror, the bottom of the mirror should be at a height of half the height of the person.

Explanation:

To see the bottom of the object in the mirror, the person should be able to see an image of the bottom of the object reflected in the mirror.

Using the law of reflection, we can determine the height h at which the bottom of the mirror should be. The angle of incidence for the light from the bottom of the object is equal to the angle of reflection, so the height of the mirror should be equal to the height of the person's eyes, H, plus the height of the bottom of the object, h. We can calculate h using similar triangles:

h/H = d/D

where d is the distance from the object to the mirror and D is the distance from the person to the mirror. Substituting the given values, we have:

h = (d/D) * H

Substituting d = D/2, we get:

h = (D/2D) * H = H/2

Therefore, the bottom of the mirror should be at a height of half the height of the person.

Other Questions
The lesser tubercle of the humerus is the insertion site for the __________ muscle. The United States form of government is solely based on that of the British/// True or False What according to you a good software design? A rhombus ????? a quadrilateral.Need help on this Enter the explicit rule for the geometric sequence. 60,12,12/5,12/25,12/125,... Can someone please help me with this physics problem? I'd really appreciate any help I can get! Why were japanese-americans placed in internment camps In Janies homeroom, 45% of the students families own a dog and 23% own a dog and a cat. What is the probability that a students family owns a cat if the family owns a dog? The Unified Coordination Group: A. Directs the incident command structure established at the incident. B. Provides coordination leadership at the Joint Field Office. C. Is a temporary Federal facility. D. Is a state-level group comprised of the heads of each Emergency Support Function (ESF) The dishes have been sorted into cups and plates. The number of plates is four less than two times the number of cups. The dishes are 60% plates. How many cups are there? a) 3b) 6c) 8d) 9 I need help with this problem. This is confusing with this formula. AT&T reacted to the popularity of the cellular phone by adding several cellular models to its line of regular phones. Availability and popularity of cellular phones is most likely due to changes in the __________________________ environments. Please help if you know the answer...if anyone can show the steps that'd be great For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency (1.21010 Hz ) and wavelength (0.026 m ). Recall that the Avogadro constant is 6.0221023 mol1. A researcher wants to observe preschoolers at a local public playground to evaluate levels of cooperation. The researcher will not interact with the children or record information in such a manner that the identity of the subjects can be readily ascertained. Which of the following statements is true? a.This research would require full committee review because it involves children. b.This research would be eligible for expedited IRB review because the researcher is not interacting with the children and the playground is public. c.This research would be eligible for exemption because the researcher is not interacting with the children and the playground is a public setting. d.This research would be eligible for expedited IRB review because the researcher is participating in the activities observed. he density of copper is 8.96g/cm^3 and the density of water is 1 g/cm^3. When a copper is submerged in a cylindrical beaker whose bottom has surface area 10 cm^2 the water level rises by 2 cm. The volume of the cylinder is the area of its base times its height. a) What is the specific gravity of copper?b) What is the buoyant force on the copper object?c) What is the buoyant force on the copper object?d) What is the mass of the copper object? What is the median of the data set given below?42, 20, 12, 15, 18, 15, 29, 33. 19. 15 Which of the following is the best example of a pioneer species?A. lichens growing on rocks after a volcanic eruptionB. grasses growing after a forest firec. oak trees replacing pine trees in a forestD. small mammals moving into a grassland You've already seen the value of 9.8 in this lesson. What's this value called? What quantity does it represent? Which of these sentences uses quotation marks correctly? Select one: a. When he saw the bird flying in the house he got angry and demanded to know, "who left the door open!" b. When he saw the bird flying in the house he got angry and demanded to know "Who Left the door open!" c. When he saw the bird flying in the house he got angry and demanded to know "who left the door open"! D. When he saw the bird flying in the house he got angry and demanded to know "who left the door open!"