If there is a potential difference V between the metal and the detector, what is the minimum energy Emin that an electron must have so that it will reach the detector

Answers

Answer 1

Answer:

Emin= eV

Explanation:

This minimum amount of energy is the work function and so for an electron to reach the detector the energy must b more than work function and the stopping potential so

Emin=eV


Related Questions

The buoyant force on an object placed in a liquid is (a) always equal to the volume of the liquid displaced. (b) always equal to the weight of the object. (c) always equal to the weight of the liquid displaced. (d) always less than the volume of the liquid displaced.

Answers

Answer:

(c) always equal to the weight of the liquid displaced.

Explanation:

Archimedes principle (also called physical law of buoyancy) states that when an object is completely or partially immersed in a fluid (liquid, e.t.c), it experiences an upthrust (or buoyant force) whose magnitude is equal to the weight of the fluid displaced by that object.

Therefore, from this principle the best option is C - always equal to the weight of the liquid displaced.

Which statement describes a disadvantage of using natural gas as an energy source? It is expensive to use. It is hard to stop using. It is a renewable resource. It is scarce in some parts of the world.

Answers

Answer:

B : It is hard to stop using.

Explanation:

just took the quiz ! hope this helps with anyone who needs it !

Due to the dependency on natural gas as a fuel, it is hard to stop using.

What is natural gas?

Natural gas is a fossil fuel which is obtained from the ground in association with petroleum.

Natural gas consists mainly of petroleum.

It is a non-renewable energy source.

Natural gas use contributes to global warming

However, due to the dependency on natural gas as a fuel, it is hard to stop using.

Learn more about natural gas at: https://brainly.com/question/815922

A ball is thrown horizontally from the top of a 41 m vertical cliff and lands 112 m from the base of the cliff. How fast is the ball thrown horizontally from the top of the cliff?

Answers

Answer:

4.78 second

Explanation:

given data

vertical cliff = 41 m

height = 112 m

solution

we know here time taken to fall vertically from the cliff =  time taken to move horizontally   ..........................1

so we use here vertical component of ball

and that is accelerated motion with initial velocity = 0

so we can solve for it as

height = 0.5 ×  g ×  t²     ........................2

put here value

112 = 0.5 ×  9.8 ×  t²    

solve it we get

t²   = 22.857

t = 4.78 second

ball thrown horizontally from the top of the cliff in 4.78 second

An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.10 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 37.5 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target

Answers

Answer:

The speed of the arrow after passing through the target is 30.1 meters per second.

Explanation:

The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:

[tex]m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}[/tex]

Where:

[tex]m_{a}[/tex], [tex]m_{t}[/tex] - Masses of arrow and target, measured in kilograms.

[tex]v_{a,o}[/tex], [tex]v_{a,f}[/tex] - Initial and final speeds of the arrow, measured in meters per second.

[tex]v_{t,o}[/tex], [tex]v_{t,f}[/tex] - Initial and final speeds of the target, measured in meters per second.

The final speed of the arrow is now cleared:

[tex]m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})[/tex]

[tex]v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})[/tex]

If [tex]v_{a,o} = 2.1\,\frac{m}{s}[/tex], [tex]m_{t} = 0.3\,kg[/tex], [tex]m_{a} = 0.0225\,kg[/tex], [tex]v_{t,o} = 2.10\,\frac{m}{s}[/tex] and [tex]v_{t,f} = 0\,\frac{m}{s}[/tex], the speed of the arrow after passing through the target is:

[tex]v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )[/tex]

[tex]v_{a,f} = 30.1\,\frac{m}{s}[/tex]

The speed of the arrow after passing through the target is 30.1 meters per second.

There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in England as a red glow on and around the Moon. How long (in s) after the asteroid hit the Moon, which is 3.77 ✕ 105 km away, would the light first arrive on Earth?

Answers

Answer:

Explanation:

speed of light = 3 x 10⁸ m /s .

distance between moon and the earth = 3.77 x 10⁵  x 10³m .

Time taken by light to cover the distance

= distance / speed

= 3.77 x 10⁸ / 3 x 10⁸

= 1.256  s

If you could see stars during the day, this is what the sky would look like at noon on a given day. The Sun is near the stars of the constellation Gemini. Near which constellation would you expect the Sun to be located at sunset?

Answers

Answer:

The sun will be located near the Gemini constellation at sunset

A box experiencing a gravitational force of 600 N. is being pulled to the right with a force of 250 NA 25 N. frictional force acts on the box as it moves to the right what is the net force in the Y direction

Answers

Answer:

32

Explanation:

Answer:

0

Explanation:.

Indiana Jones is in a temple searching for artifacts. He finds a gold sphere with a radius of 2 cm sitting on a pressure sensitive plate. To avoid triggering the pressure plate, he must replace the gold with something of equal mass. The density of gold is 19.3.103 kg/m3, and the volume of a sphere is V = 4/3 Ar3. Indy has a bag of sand with a density of 1,602 kg/m3.
(A) What volume of sand must he replace the gold sphere with? If the sand was a sphere, what radius would it have?

Answers

Answer:

Volume of Sand = 0.4 m³

Radius of Sand Sphere = 0.46 m

Explanation:

First we need to find the volume of gold sphere:

Vg = (4/3)πr³

where,

Vg = Volume of gold sphere = ?

r = radius of gold sphere = 2 cm = 0.02 m

Therefore,

Vg = (4/3)π(0.2 m)³

Vg = 0.0335 m³

Now, we find mass of the gold:

ρg = mg/Vg

where,

ρg = density of gold = 19300 kg/m³

mg = mass of gold = ?

Vg = Volume of gold sphere = 0.0335 m³

Therefore,

mg = (19300 kg/m³)(0.0335 m³)

mg = 646.75 kg

Now, the volume of sand required for equivalent mass of gold, will be given by:

ρs = mg/Vs

where,

ρs = density of sand = 1602 kg/m³

mg = mass of gold = 646.75 kg

Vs = Volume of sand = ?

Therefore,

1602 kg/m³ = 646.75 kg/Vs

Vs = (646.75 kg)/(1602 kg/m³)

Vs = 0.4 m³

Now, for the radius of sand sphere to give a volume of 0.4 m³, can be determined from the formula:

Vs = (4/3)πr³

0.4 m³ = (4/3)πr³

r³ = 3(0.4 m³)/4π

r³ = 0.095 m³

r = ∛(0.095 m³)

r = 0.46 m

A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a stationary particle with a charge of –5 × 10–6 C. The radius of the orbit is:

Answers

Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

[tex]F_c=F_e=ma_c[/tex]      (1)

m: mass of the particle = 20g = 20*10-3 kg

ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

[tex]F_e=k\frac{qq'}{r^2}[/tex]             (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

[tex]a_c=\frac{v^2}{r}[/tex]                 (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

[tex]k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}[/tex]

Finally, you replace the values of all parameters in the previous expression:

[tex]r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m[/tex]

The radius of the circular trajectory is 0.22m

Calculate the moment of inertia of a skater given the following information.
(a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius.
(b) The skater with arms extended is approximately a cylinder that is 74.0 kg, has a 0.150 m radius, and has two 0.750 m long arms which are 3.00 kg each and extend straight out from the cylinder like rods rotated about their ends.

Answers

Answer:

(a)  I = 0.363 kgm^2

(b)  I = 1.95 kgm^2

Explanation:

(a) If you consider the shape of the skater as approximately a cylinder, you use the following formula to calculate the moment of inertia of the skater:

[tex]I_s=\frac{1}{2}MR^2[/tex]         (1)

M: mass of the skater = 60.0 kg

R: radius of the cylinder = 0.110m

[tex]I_s=\frac{1}{2}(60.0kg)(0.110m)^2=0.363kg.m^2[/tex]

The moment of inertia of the skater is 0.363 kgm^2

(b) In the case of the skater with his arms extended, you calculate the moment of inertia of a combine object, given by cylinder and a rod (the arms) that cross the cylinder. You use the following formula for the total moment of inertia:

[tex]I=I_c+I_r\\\\I=\frac{1}{2}M_1R^2+\frac{1}{12}M_2L^2[/tex]       (2)

M1: mass of the cylinder = 74.0 kg

M2: mass of the rod = 3.00kg +3.00kg = 6.00kg

L: length of the rod = 0.750m + 0.750m = 1.50m

R: radius of the cylinder = 0.150

[tex]I=\frac{1}{2}(74.0kg)(0.150m)^2+\frac{1}{12}(6.00kg)(1.50m)^2\\\\I=1.95kg.m^2[/tex]

The moment of inertia of the skater with his arms extended is 1.95 kg.m^2

A very long, straight horizontal wire carries a current such that 8.25×1018 electrons per second pass any given point going from west to east.
What is the magnitude of the magnetic field this wire produces at a point 5.00 cm directly above it?

Answers

Answer: The magnitude of the magnetic field this wire produces is 5.56 × 10^-6 T

Explanation: Please see the attachments below

This question involves the concepts of Biot-Savart Law and magnetic field. It can be solved by the application of Biot-Savart Law on a current-carrying wire.

The magnitude of the magnetic field is "5.28 x 10⁻⁶ T".

Using the Biot-Savart Law to find out the magnitude of the magnetic field produced by the wire at a point 5 cm above it, directly:

[tex]B = \frac{\mu_o I}{2\pi r}\\\\[/tex]

where,

B = magnetic field = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = [tex]\frac{ne}{t} = (8.25\ x\ 10^{18}\ elctrons/sec)(1.6\ x\ 10^{-19}\ C /electron)[/tex] = 1.32 A

r = radius = distance above wire = 5 cm = 0.05 m

Therefore,

[tex]B = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(1.32\ A)}{2\pi(0.05\ m)}[/tex]

B = 5.28 x 10⁻⁶ T

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The attached picture shows the illustration of magnetic field due to a wire.

g A point mass of 1.5kg is attached to a spring and set to oscillate through simple harmonic oscillations. If the period of the oscillation is 10s, find the spring constant.

Answers

Answer:

k = 0.6 N/m

Explanation:

The time period of a spring mass oscillation system is given by the following formula:

T = 2π√(m/k)

where,

T = Time Period of Oscillation = 10 s

m = Mass attached to the spring = 1.5 kg

k = spring constant = ?

Therefore,

10 s = 2π√(1.5 kg/k)

squaring on both sides we get:

100 s² = 4π²(1.5 kg/k)

k = 6π² kg/100 s²

k = 0.6 N/m

An electric heater is constructed by applying a potential different of 120V across a nichrome wire that has a total resistant of 8 ohm .the current by the wire is

Answers

Answer:

15amps

Explanation:

V=IR

I=V/R

I = 120/8

I = 15 amps

The cost of energy delivered to residences by electrical transmission varies from $0.070/kWh to $0.258/kWh throughout the United States; $0.110/kWh is the average value.

Required:
At this average price, calculate the cost of:

a. leaving a 40-W porch light on for two weeks while you are on vacation?
b. making a piece of dark toast in 3.00 min with a 970-W toaster
c. drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer.

Answers

Answer:

Cost = $ 1.48

Cost = $ 0.005

Cost = $ 0.38

Explanation:

given data

electrical transmission varies = $0.070/kWh to $0.258/kWh

average value = $0.110/kWh

solution

when leaving a 40-W porch light on for two weeks while you are on vacation so cost will be

first we get here energy consumed that is express as

E = Pt    .................1

here E is Energy Consumed and Power Delivered is P and t is time

so power is here 0.04 KW and t = 2 week = 336 hour

so

put value in 1 we get

E = 0.04 × 336

E = 13.44 KWh

so cost will be as

Cost = E × Unit Price    .............2

put here value and we get

Cost = 13.44 × 0.11

Cost = $ 1.48

and

when you making a piece of dark toast in 3.00 min with a 970-W toaster

so energy consumed will be by equation 1 we get

E = Pt

power is = 0.97 KW and time = 3 min = 0.05 hour

put value in equation 1 for energy consume

E = 0.97 × 0.05 h

E = 0.0485 KWh

and we get cost by w\put value in equation 2 that will be

cost =  E × Unit Price

cost = 0.0485 × 0.11

Cost = $ 0.005

and

when drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer

from equation 1 we get energy consume

E = Pt

Power Delivered = 5.203 KW and time = 40 min = 0.67 hour

E = 5.203 × 0.67

E = 3.47 KWh

and

cost will by put value in equation 2

Cost = E × Unit Price

Cost = 3.47 × 0.11

Cost = $ 0.38

Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.160. If the patch is of width 62.0 m and the average force of air resistance on the skier is 160 N , how fast is she going after crossing the patch?

Answers

Answer:

14.1 m/s

Explanation:

From the question,

μk = a/g...................... Equation 1

Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.

make a the subject of the equation

a = μk(g).................. Equation 2

Given: μk = 0.160, g = 9.8 m/s²

Substitute into equation 2

a = 0.16(9.8)

a = 1.568 m/s²

Using,

F = ma

Where F = force, m = mass.

Make m the subject of the equation

m = F/a................... Equation 3

m = 160/1.568

m = 102.04 kg.

Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.

Assuming the initial velocity of the skier to be zero

Fd+mgμ = 1/2mv²........................Equation 4

Where v = speed of the skier after crossing the patch, d = distance/width of the patch.

v = √2(Fd+mgμ)/m)................ Equation 5

Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16

Substitute these values into equation 5

v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]

v = √197.57

v = 14.1 m/s

v = 9.86 m/s

A student is conducting an experiment that involves adding hydrochloric acid to various minerals to detect if they have carbonates in them. The student holds a mineral up and adds hydrochloric acid to it. The acid runs down the side and onto the student’s hand causing irritation and a minor burn. If they had done a risk assessment first, how would this situation be different? A. It would be the same, there is no way to predict the random chance of acid dripping off the mineral in a risk assessment. B. The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. C. The student would be safer because he would have been wearing goggles, but his hand still would not have been protected. D. The student would not have picked up the mineral because he would know that some of the minerals have dangerous chemicals in them.

Answers

The answer would be D because it could have been prevented

By  the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "

What is experiment ?

An experiment would be a technique used to confirm or deny a hypothesis, as well as assess the likelihood or effectiveness of something that has never been tried before.

What is hydrochloric acid?

Hydrochloric acid is a kind of compound in which hydrogen and chlorine element is present.

Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.

By  the experiment "By  the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "

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A certain dam generates 120 MJ of mechanical (hydroelectric) energy each minute. If the conversion from mechanical to electrical energy is then 15% efficient, what is the dam's electrical power output in W?

Answers

Answer:

electric energy ( power ) = 300000 W

Explanation:

given data

mechanical (hydroelectric) energy = 120 MJ/min = 2000000 J/s

efficiency = 15 % = 0.15

solution

we know that Efficiency of electric engine is expression  as

Efficiency = Mechanical energy ÷ electric energy  ......................1

and here dam electrical power output is

put here value in equation 1

electric energy ( power ) = Efficiency × Mechanical energy ( power )

electric energy ( power ) = 0.15 × 2000000 J/s

electric energy ( power ) = 300000 W

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to

Answers

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is  [tex]v = 3.79 *10^{5} \ m/s[/tex]  

Explanation:

From the question we are told that

    The  magnitude of the electric field is  [tex]E = 144 \ kV /m = 144*10^{3} \ V/m[/tex]

     The magnetic field is  [tex]B = 0.38 \ T[/tex]

   

The force due to the electric field is mathematically represented as

      [tex]F_e = E * q[/tex]

and

The force due to the magnetic field is mathematically represented as

    [tex]F_b = q * v * B * sin(\theta )[/tex]

Now given that it is perpendicular ,  [tex]\theta = 90[/tex]

=>   [tex]F_b = q * v * B * sin(90)[/tex]

=>   [tex]F_b = q * v * B[/tex]

Now  given that it is not deflected it means that

        [tex]F_ e = F_b[/tex]

=>    [tex]q * E = q * v * B[/tex]

=>   [tex]v = \frac{E}{B }[/tex]

 substituting values

     [tex]v = \frac{ 144 *10^{3}}{0.38 }[/tex]

     [tex]v = 3.79 *10^{5} \ m/s[/tex]

A heavier car is always safer in a crash than a lighter car.

Answers

Answer:

not true because the mass from the heavy car will cause it to damage more

Explanation:

Answer: answer on edmentum is false your welcome

Explanation:

It is false because it's more heavy so more damage l.

The friends spend some time thinking about a beam of light traveling from one medium to another medium with higher index of refraction, which strikes the boundary obliquely. Which of Tristan's statements is correct

Answers

Answer:

"When light moves from a material in which its speed is high to a material in which its speed is lower, the angle of refraction θ2is less than the angle of incidence θ1and the ray is bent toward the normal."

Explanation:

Refraction is a phenomenon that occurs when light rays change direction after passing through a surface or medium. This is also known as 'bending'. Snell's law provides the relationship between the angle of incidence and refraction in the equation below:

n₁sinФ₁ = n₂sinФ₂

where n1 and n2 represent the two media and theta refers to the angles formed. When light hits a medium with a high refractive index, the speed of light becomes slower.

So, Tristan is right when he says that, "When light moves from a material in which its speed is high to a material in which its speed is lower, the angle of refraction θ2is less than the angle of incidence θ1and the ray is bent toward the normal."

Find the length (in m) of an organ pipe closed at one end that produces a fundamental frequency of 175 Hz when air temperature is 18.0°C

Answers

Answer:

Length = 0.4882 m

Explanation:

given data

fundamental frequency = 175 Hz

air temperature = 18.0°C

solution

we will apply here fundamental frequency formula that is

F = [tex]\frac{v}{4L}[/tex]      ....................1

here v =   [tex]331 \sqrt{1+\frac{T}{273}}[/tex]  

here 331 m/s is speed of sound in air

so v =  [tex]331 \sqrt{1+\frac{18}{273}}[/tex]   = 341.74 m/s

now put value in equation 1 we get

F = [tex]\frac{v}{4L}[/tex]

[tex]175 = \frac{341.74}{4L}[/tex]  

Length = 0.4882 m

Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.0° above the horizontal, and the second arrow is fired straight upward. Assume an isolated system and choose the reference configuration at the initial position of the arrows.
(a) what is the maximum height of each of the arrows?
(b) What is the total mechanical energy of the arrow-Earth system for each of the arrows at their maximum height?

Answers

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

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

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{x,1}[/tex], [tex]K_{x,2}[/tex] - Initial and final horizontal translational kinetic energy, measured in joules.

[tex]K_{y,1}[/tex], [tex]K_{y,2}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

[tex]m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0[/tex]

[tex]g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})[/tex]

[tex]y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}[/tex]

Where:

[tex]y_{1}[/tex]. [tex]y_{2}[/tex] - Initial and final height of the arrow, measured in meters.

[tex]v_{y,1}[/tex], [tex]v_{y,2}[/tex] - Initial and final vertical speed of the arrow, measured in meters.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

The initial vertical speed of the arrow is:

[tex]v_{y,1} = v_{1}\cdot \sin \theta[/tex]

Where:

[tex]v_{1}[/tex] - Magnitude of the initial velocity, measured in meters per second.

[tex]\theta[/tex] - Initial angle, measured in sexagesimal degrees.

If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the initial vertical speed is:

[tex]v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}[/tex]

[tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex]

If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex] and [tex]v_{y,2} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:

[tex]y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]y_{2} - y_{1} = 56.712\,m[/tex]

Second arrow

[tex]U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,3}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{y,1}[/tex], [tex]K_{y,3}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.

[tex]m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0[/tex]

[tex]g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})[/tex]

[tex]y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}[/tex]

If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} = 82\,\frac{m}{s}[/tex] and [tex]v_{y,3} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:

[tex]y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]y_{3} - y_{1} = 342.816\,m[/tex]

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

[tex]E = U + K_{x}[/tex]

The expression is now expanded:

[tex]E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}[/tex]

Where [tex]v_{x}[/tex] is the horizontal speed of the arrow, measured in meters per second.

[tex]v_{x} = v_{1}\cdot \cos \theta[/tex]

If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the horizontal speed is:

[tex]v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}[/tex]

[tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex]

If [tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{max} = 56.712\,m[/tex] and [tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex], the total mechanical energy is:

[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}[/tex]

[tex]E = 201.720\,J[/tex]

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

[tex]E = m\cdot g \cdot y_{max}[/tex]

[tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{max} = 342.816\,m[/tex]

[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)[/tex]

[tex]E = 201.720\,J[/tex]

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Two radio antennas A and B radiate in phase. Antenna B is 120 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 40 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied.
(a) What is the longest wavelength for which there will be destructive interference at point Q?
(b) What is the longest wavelength for which there will be constructive interference at point Q?

Answers

Answer:

a. for destructive interference

λmax= 240m

b. for constructive interference

λmax = 120m

Explanation:

A block attached to a spring undergoes simple harmonic motion on a horizontal frictionless surface. Its total energy is 50 J. When the displacement is half the amplitude, the kinetic energy is

Answers

Answer:

The kinetic energy at a displacement of half the amplitude is 37.5 J

Explanation:

Given;

total energy on the spring, E = 50 J

When the displacement is half the amplitude, the total energy in the spring is sum of the kinetic energy and elastic potential energy.

E = K + U

Where;

K is the kinetic energy

U is the elastic potential energy

K = E - U

K = E - ¹/₂KA²

When the displacement is half = ¹/₂(A) = A/₂

K = E - ¹/₂K(A/₂)²

K = E - ¹/₂K(A²/₄)

K = E - ¹₄(¹/₂KA²)

Recall, E = ¹/₂KA²

K = ¹/₂KA² - ¹₄(¹/₂KA²)     (recall from simple arithmetic, 1 - ¹/₄ = ³/₄)

K = 1(¹/₂KA²) - ¹₄(¹/₂KA²)  = ³/₄(¹/₂KA²)

K = ³/₄(¹/₂KA²)

But E = ¹/₂KA² = 50J

K = ³/₄ (50J)

K = 37.5 J

Therefore, the kinetic energy at a displacement of half the amplitude is 37.5 J

The kinetic energy when the displacement is half the amplitude

Given the following data:

Total energy = 50 Joules.Displacement, x = [tex]\frac{A}{2}[/tex]

To find the kinetic energy when the displacement is half the amplitude:

The total energy of the system of a block and a spring is the sum of the spring's elastic potential energy and kinetic energy of the block and it's proportional to the square of the amplitude.

Mathematically, the total energy of the system of a block and a spring is given by the formula:

[tex]T.E = U + K.E[/tex]   .....equation 1.

[tex]T.E = \frac{1}{2} kA^2[/tex]

Where:

T.E is the total energy.U is the elastic potential energy.K.E is the kinetic energy.A is the amplitude.

Making K.E the subject of formula, we have:

[tex]K.E = T.E - U[/tex]   .....equation 2.

But, [tex]U = \frac{1}{2} kx^2[/tex]    ....equation 3.

Where:

k is spring constant.x is change in position (displacement).

Substituting the eqn 3 into eqn 2, we have:

[tex]K.E = T.E - \frac{1}{2} kx^2[/tex]

[tex]K.E = T.E - \frac{1}{2} k(\frac{A}{2})^2\\\\K.E = T.E - \frac{1}{2} k(\frac{A^2}{4})\\\\K.E = T.E - \frac{1}{4} (\frac{1}{2} kA^2)\\\\K.E = T.E - \frac{1}{4} (T.E)\\\\K.E = 50 - \frac{1}{4} (50)\\\\K.E = 50 - 12.5[/tex]

K.E = 37.5 Joules.

Read more: https://brainly.com/question/23153766

When we describe electric flux, we say that a surface is oriented in a certain direction with respect to an electric field. When we try to calculate how much electric field passes through the surface, we make use of the:_________.
1. Wedge Product
2. Dot Product
3. Cross Product

Answers

Answer:

2. Dot Product

Explanation:

The calculation of the electric flux gives an scalar result.

When we tray to calculate how much electric field passes trough a surface, we are calculating a scalar value. Furthermore, the concept of flux requires the calculation of a scalar value.

Also it is necessary to take into account that the magnitude of the flux trough a surface depends of the inclination of the surface respect to the direction of the electric field. This is taken into account sufficiently by a dot product.

Then, the answer is:

2. Dot Product

If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 55.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answers

Answer:

14.79 kgm/s

Explanation:

Data provided in the question

Let us assume the mass of baseball =  m = 0.145 kg

The Initial velocity of pitched ball = [tex]v_i[/tex] = 47 m/s

Final velocity of batted ball in the opposite direction = [tex]v_f[/tex]= -55m/s

Based on the above information, the change in momentum is

[tex]\Delta P = m(v_f -v_i)[/tex]

[tex]= 0.145 kg(-55m/s - 47m/s)[/tex]    

= 14.79 kgm/s

Hence, the magnitude of the change in momentum of the ball is 14.79 kg m/s

Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.

Answers

Answer:

Explanation:

Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.

When you are told that the wind has a "Small Coriolis force" associated with it, what is that "small force" exactly

Answers

Answer:

Coriolis force is a type of force of inertia that acts on objects that is in motion within a frame of reference that rotates with respect to an inertial frame. Due to the rotation of the earth, circulating air is deflected result of the Coriolis force, instead of the air circulating between the earth poles and the equator in a straight manner. Because of the effect of the Coriolis force,  air movement deflects toward the right in the Northern Hemisphere and toward the left in the Southern Hemisphere, eventually taking a curved path of travel.

help me
Describe the different types of non contact forces.​

Answers

Answer:

the correct answer is

All four known fundamental interactions are non-contact forces: Gravity, the force of attraction that exists among all bodies that have mass. ... Examples of this force include: electricity, magnetism, radio waves, microwaves, infrared, visible light, X-rays and gamma rays.

Explanation:

hope this helps you!!!!

A uniform 2.0-kg rod that is 0.92 m long is suspended at rest from the ceiling by two springs, one at each end. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 29 N/m and 66 N/m. Find the angle that the rod makes with the horizontal.

Answers

Answer:

11.7°

Explanation:

See attached file

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