An ideal gas is enclosed in a cylinder with a movable piston on top of it. The piston has a mass of 8,000 g and an area of 5.00 cm2 and is free to slide up and down, keeping the pressure of the gas constant.
a. How much work is done on the gas as the temperature of 0.280 mol of the gas is raised from 15.0 degrees C to 27.0 degrees C?
b. What does the sign of your answer to part (a) indicate?
A. The surroundings do positive work on the gas.
B. The gas does positive work on its surroundings.
C. There is no work done by the gas of the surroundings.

Answer:

λ = 157 μm

Explanation:

given,

[tex]\vec{E}(y,t)=[ 3.9\times 10^5\ V/m cos(ky-(1.20\times 10^{13}\ rad/s )t)]\vec{k}[/tex]

now,

ω = 1.2 x 10¹³ rad/s

[tex]f =\dfrac{\omega}{2\pi}[/tex]

[tex]f =\dfrac{1.2\times 10^{13}}{2\pi}[/tex]

f = 1.91 x 10¹² Hz

determine the wavelength of the E M wave

 [tex]\lambda = \dfrac{c}{f}[/tex]

 [tex]\lambda = \dfrac{3\times 10^8}{1.91\times 10^{12}}[/tex]

   λ = 1.57 x 10⁻⁴ m

   λ = 157 x 10⁻⁶ m

   λ = 157 μm

hence, the wavelength of the wave is equal to λ = 157 μm

Given that,

here,

As we know,

→ [tex]f = \frac{\omega}{2 \pi}[/tex]

By substituting the values, we get

     [tex]= \frac{1.2\times 10 ^{13}}{2 \pi}[/tex]

     [tex]= 1.91\times 10^{12} \ Hz[/tex]

hence,

The wavelength will be:

→ [tex]\lambda = \frac{c}{f}[/tex]

     [tex]= \frac{3\times 10^8}{1.91\times 10^{12}}[/tex]

     [tex]=157\times 10^{-6} \ m[/tex]

     [tex]= 157 \ \mu m[/tex]

Thus the above response is correct.

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Balanced forces are equal and opposite, causing no change in an object's movement, leading to static equilibrium. On the contrary, unbalanced forces change an object's velocity or direction, causing it to move or accelerate.

The primary difference between balanced and unbalanced forces lies in their effects on the motion of an object. Balanced forces are equal and opposite forces that act on an object, resulting in no change in its velocity or direction. An example would be a static car or an ice hockey stick lying flat with two equal and opposite forces applied to it. Friction is negligible and the gravitational force is balanced by the support, leading to static equilibrium.

On the other hand, unbalanced forces are not equaled.

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Answer:

Technician B

Explanation:

Car engine compression refers to when air and gas are mixed together in the cylinders of an engine. This process is required for the car to move and function. If there are any problems with the compression process, then you can expect to experience all kinds of car problems.

It will be easy to tell when you have a low compression problem because you may experience a misfire when you try to start the engine. Either that or the engine will offer poor performance as you’re driving the vehicle down the road. The worst-case scenario would be the car not starting if all the cylinders have no compression.

Generally speaking, if you have low compression in one(single) cylinder, the engine will start but you’ll likely experience misfires and your vehicle will run rough. If you experience no compression in ALL cylinders, your engine simply won’t start. This rule out A statement.

An engine tune-up is a exercise for engines to undergo regularly. It is a way of making a car's work at the level and standard intended by the car manufacturer when the car was first made. All manufacturers will stipulate a schedule of when a car will required an engine tune-up to ensure that a car runs at it's most efficient

Engine tune-up are imperative to ensure that all the power and efficiency that  your car is capable are being reached . This bring the answer to the question to Tech B.

Answer:

option B

Explanation:

given,

spring constant = k = 100 N/m

distance of unstretched length = 0.5 m

distance after stretched length =  0.65 m

magnitude of weight = ?

We know,

W  = K x

x is the displacement

x = 0.65 - 0.5 = 0.15 m

W = K x

W  = 100 x 0.15

W  = 15 N

Hence, the correct answer is option B

Final answer:

The weight of the attached object is found by multiplying the spring constant (100 N/m) by the stretch in the spring (0.15 m), which equals 15 N. Therefore, the weight of the object is 15 N.

Explanation:

The weight of the attached object can be calculated using Hooke's Law, which states that the force exerted by the spring (F) is equal to the spring constant (k) multiplied by the displacement from the rest position ( extDelta x): F = k extDelta x. In this case, the spring has a spring constant of 100 newtons per meter and the displacement from the original length of 0.50 meters to the new length of 0.65 meters is 0.15 meters.

Thus, F = 100 N/m exttimes 0.15 m, which gives us a force of 15 N. Since the weight of the object is the force exerted by the spring and these two forces are equal and opposite when the spring is at rest with the object hanging, the weight of the attached object is 15 N. Therefore, the correct answer is (b) 15N.

Answer:

Utilize the Private VLAN feature to segregate network traffic at Layer 2, and

Utilize the ARP inspection feature to help prevent the misuse of gARP.

Explanation:

An Engineer designing layer 2 and layer 3 network, must Utilize the Private VLAN feature to segregate network traffic at Layer 2 and Utilize the ARP inspection feature to help prevent the misuse of gARP.

Answer:

Stooping more Quickly

Explanation: Stooping more quickly means that Motor cycles can easily bend and stop,this is due to it's design, Motor cycle have only two wheels,they contain just two tyres one in front and one at the back,they are not as firm on the ground when compared to other vehicles such Motor car. They have a lower weight which makes them easily bent or stopped once there's any barrier.

Motorcycles can stop quicker than other vehicles.

When following a motorcycle, drivers must remember that motorcycles can stop quicker than other vehicles. Due to their smaller size and weight, motorcycles have a shorter stopping distance. This means that drivers should maintain a safe following distance to avoid rear-ending the motorcycle.

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Answer:

C 30 mph

Explanation:

The speed limits in various area's can be concluded as

1. 25- 30 mph in urban residential areas and school districts.

2.55 mph on rural highways, and

3. 70 mph on rural Interstate highways.

so, the correct answer is c

If the initial speed of the box at Point A is increased by a factor of 2, the final velocity, denoted as W, will also increase by a factor of 2.

If the initial speed of the box at Point A is increased by a factor of 2, the final velocity of the box, denoted as W, will also increase by a factor of 2.

For example, if the initial speed of the box was 10 m/s, after increasing it by a factor of 2, the new speed would be 20 m/s.

Therefore, option (a), It will increase by a factor of two, is the correct answer choice.

Answer:

a= 2.667 m/s²

Explanation:

Given that

Initial velocity ,u= 2 m/s

Final velocity ,v= 10 m/s

Time taken ,t= 3 s

As we know that

v =  u +  at

a=Acceleration ,u=Initial velocity ,v= Final velocity ,t=time

Now by putting the values in the above equation

v =  u +  at

10 = 2 + a x 3

8 = 3 a

[tex]a=\dfrac{8}{3}\ m/s^2[/tex]

a= 2.667 m/s²

Therefore the acceleration of the biker will be 2.667 m/s²

Answer:

The angular velocity when the arms are pulled is ω₂= 6.25 rad/s

Explanation:

Assuming that you want to determine the final angular velocity when is the arms are pulled, then it is calculated using the principle of conservation of angular momentum. It states that:

I₁ω₁=I₂ω₂

where I = moment of inertia , ω= angular velocity , 1 and 2 denote the skater with extended hands and pulled respectively.

Thus

I₁ω₁=I₂ω₂

ω₂= I₁ω₁/I₂

replacing values

ω₂= ω₁ *(I₁/I₂) = 5 rad/s *(2.25 kg·m2/1.80 kg·m2) = 6.25 rad/s

ω₂= 6.25 rad/s

Answer:

Explanation:

Laws of  refraction:

(a) The incident ray, the refracted ray and the normal at the point of incident all lies in the same plane

(b) The ratio of the sine of incident  to the sine of refraction is a constant for a given pair of media, which is the refractive index of the second medium with respect to the first medium. The is also called Snell's law

from Snell's law,

Index of refraction of the glass = sini/sinr.............. Equation 1

where i = incident angle, r = angle of refraction.

given: i = 45°, r = 30°

Substituting these values into equation 1,

Index of refraction of the glass = sin45°/sin30°

Index of refraction of the glass = (1/√2)/(1/2)

Index of refraction of the glass = 2/√2

Index of refraction of the glass = √2.

Therefore, The Index of refraction of the glass = √2.

Final answer:

To calculate the index of refraction of the glass, we apply Snell's Law, resulting in an index of refraction of approximately 1.41.

Explanation:

To find the index of refraction of the glass, we can use Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the two media. Snell's Law states:

n1 * sin(θ1) = n2 * sin(θ2),

where n1 and n2 are the indices of refraction of the first and second media (air and glass, respectively), and θ1 and θ2 are the angles of incidence and refraction. Given that the index of refraction of air is approximately 1 (since it's very close to a vacuum) and using the provided angles (45° and 30°), the calculation will look like this:

1 * sin(45°) = n2 * sin(30°)

Which simplifies to:

1 * √2/2 = n2 * 1/2

After calculating, we find:

n2 = √2

Thus, the index of refraction of the glass is √2, which is approximately 1.41.

Answer:

C.inertia

Explanation:

Inertia is a property of matter to maintain its state of motion.

In this case, if the car was moving along with the people in it when the car stopped, the inertia of the people will cause them to continue moving.

This is why in a crash a person can be shot from the car, due to the inertia he has from his previous state of motion.

The complete sentence is: Unrestrained occupants of a car keep moving during the time the car takes to stop due to inertia. (option C)

Unrestrained occupants of a car continue moving forward when the car stops due to inertia, as an object in motion remains in motion unless acted upon by an external force.

Unrestrained occupants of a car keep moving during the time the car takes to stop due to inertia. This phenomenon occurs because, according to Newton's first law, an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. In the context of a car, when it suddenly stops, the passengers inside will continue moving forward because their bodies were initially in motion and there is no forward force directly applied to them to cause a deceleration similar to that of the car.

Centripetal force is the force that causes an object moving in a circular path to move towards the center of the path. However, when the force disappears, inertia is what carries the object in a straight line, tangent to the circular path. While centrifugal force is often used in common language to describe the sensation of being pushed outward in a rotating system, it is not an actual force but rather a perceived effect due to inertia.

Answer:

  v₂ = 2 v₁   blood velocity in the built zone is twice the speed of the normal zone,   so the correct response is c

Explanation:

For that problem we must use the fluid continuity equation

          A₁ v₁ = A₂ v₂

Where point 1 is in the normal artery and point 2 in the contraction

         v₂ = v₁  A₁ / A₂

In the exercise indicates

          A₂ = ½ A₁

          v₂ = 2 v₁

The blood velocity in the built zone is twice the speed of the normal zone, so the correct response is c

Answer:

Ķ rot = 2K rot

When the arms and weight are retracted toward the body, the energy increases by twice the initial kinetic energy because the moment of inertia has decreased coupled with the energy needed to bring the hands towards the body.

Explanation:

K rot = kinetic energy of system when hands + weight are stretched out = ½ Iw²

Where w = angular velocity when arms and weight are stretched out

I = inertia of the system = ( Ie + Ip) where Ie= moment of inertia of the extra weight carried and Ip = moment of inertia of the person.

Ķ rot = final rotational energy when arms and external weight are pulled in = ½Ìŵ² where Ì =( Ìe + Ìp)

And Ìe = inertia of weight when hands are retracted and Ìp = inertia of person when hand are retracted

Using conservation of angular momentum which is

Iw = Ìŵ

Substitute ŵ = (I/Ì)*W in Ķ rot to give

Ķ rot =½Ì *[ (I/Ì) * W ]²

Ķ rot = ½ ( I²/Ì) * w²

Since K rot = ½ I w²

So Ķ rot = K rot ( I/Ì )...eq3

From the question above system inertia reduces by a factor of 2

Ì = ½I

Sub expression into equ3

Ķ rot = 2K rot

Answer: 14.62 Earth days

Explanation:

This problem can be solved by Kepler’s Third Law of Planetary motion:

[tex]T=2 \pi \sqrt{\frac{a^{3}}{GM}}[/tex]

Where:

[tex]T[/tex] is the period of the probe

[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Gravitational Constant

[tex]M=1(10)^{13} kg[/tex] is the mass of the comet Churyumov–Gerasimenko

[tex]a=30 km \frac{1000 m}{1 km}=30000 m[/tex] is the semimajor axis of the orbit the probe described around the comet (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

[tex]T=2 \pi \sqrt{\frac{(30000 m)^{3}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(1(10)^{13} kg)}}[/tex]

[tex]T=1,263,771.768 s \frac{1 h}{3600 s} \frac{1 Earth-day}{24 h}=14.62 Earth-days[/tex]

Hence, the orbital period of the probe is 14.62 Earth days.

The orbital period of the Rosetta space probe, in orbit around the comet Churyumov–Gerasimenko in 2014, is calculated to be approximately 0.178 Earth days using Kepler's Third Law of Planetary Motion and given values.

This question pertains to the calculation of the orbital period of the Rosetta space probe when it entered orbit around the comet Churyumov–Gerasimenko in 2014. First, we establish the gravitational constant (G) as approximately 6.674 × 10^-11 m^3 kg^-1 s^-2. To calculate the orbital period of an object in orbit around a celestial body, we can use Kepler's Third Law of Planetary Motion which states the square of the orbital period T is proportional to the cube of the semi-major axis a. Rearranging this law and substituting, we get T = 2π sqrt(a^3/(G*M)). Applying the given values, we get T = 2π sqrt((30*10^3 m)^3/(6.674 × 10^-11 m^3 kg^-1 s^-2*1.0*10^13 kg)) = 15378 seconds or approximately 0.178 days.

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Explanationhe rotational kinetic energy is

[tex]K_{r} =\frac{1}{2} Iω^{2}[/tex]

The moment of inertia  I  for a sphere is  ( 2 / 5 ) m r ^2

. Substituting this in the equation yields

Kr=1/2( ( 2 / 5 ) m r ^2 )([tex](\frac{v}{r})^{2}[/tex]

1/5mv^2

1/5*5.97 × 10 ^24 *(2[tex]\pi[/tex]*6.38*10^6/86400)^2

2.57 × 10 ^29 J

b. kinetic energy of the sun

K.E=1/2*mv^2

the distance from the earth to the sun is given as

.

Answer:

a. 7.43 × 10³⁴ J b. 3.51 × 10³⁸ J

Explanation:

a. The gravitational force of attraction of a body on the surface of the earth equals the centripetal force on it due to the earth.

So, GMm/R² = mRω²

ω = √(GM/R³) where ω = angular speed of the earth. M = mass of earth = 5.972 × 10²⁴ kg, R = radius of earth = 6.4 × 10⁶ m and G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

The rotational kinetic energy of earth K.E = 1/2Iω² where I = rotational inertia = 2/5MR²

K.E = 1/2Iω²

= 1/2 × 2/5MR² × GM/R³

= GM²/5R

= 6.67 × 10⁻¹¹ Nm²/kg² × (5.972 × 10²⁴ kg)² /(6.4 × 10⁶ m × 5)

= 7.43 × 10³⁴ J

b. Similarly, the rotational kinetic energy of the earth around the sun is

K.E = GM²/5R where M = mass of sun = 1.989 × 10³⁰ kg and R = distance of earth from sun = 1.5047 × 10¹¹ m

K.E = GM²/5R

= 6.67 × 10⁻¹¹ Nm²/kg² × (1.989 × 10³⁰ kg)² / (1.5047 × 10¹¹ m × 5)

= 3.5073 × 10³⁸ J ≅ 3.51 × 10³⁸ J

Answer:

Resistance of order of 10Kiloohms

Explanation:

Pull up resistors are essential in every microcontroller devices or digital circuits. Which helps prevent floating ( a process of not being able to determine the state of a pin) It works by allowing the state of a pin to be in either a high state or low state. A resistor is connected between the Vcc and the pin.

When the switch is open, the entire voltage flows through the resistor (V=IR). When the switch is closed the pull up resistor and the internal impedance of the pin form a voltage divider to limit the current flow.

The maximum value of a pull-up resistor to avoid floating when the switch is open is typically between 1KΩ to 10KΩ. This helps to prevent excessive current flow and quickly pulls the input pin high to avoid a floating state. Extremely high resistance values, like 1 megaohm, may cause brief floating, potentially leading to incorrect readings.

The maximum value of the pull-up resistor, in order to avoid a floating condition when the switch is open, depends on several factors including the characteristics of the microcontroller's input pin and the voltage level of the system. However, a moderately high resistance value is often chosen in the range of 1KΩ to 10KΩ to prevent excessive current flow when the switch is closed while still pulling the input pin high quickly enough to avoid a floating state when the switch is open. Remember that a pull-up resistor's purpose is to define a state of a digital input pin when it is not otherwise driven by an external component or system.

In some cases, the maximum value of the pull-up resistor will need to be calculated based on the specifications of your particular circuit. An extremely high resistance, like 1 megaohm, can cause the input to float for a very short time, long enough that your microcontroller might read it incorrectly.

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Answer:

A. 200 J

Explanation:

The initial kinetic energy depends on the initial speed, while the gravitational potential energy depends on the height, both balls are thrown with the same initial speed and from the same height. Therefore, due to the law of conservation of energy, the balls must have the same mechanical energy (the sum of both energies) when both impact the ground. Since the potential energy is zero at this point, its final kinetic energy must also be the same.

The second ball, thrown vertically upward, will have kinetic energy of 200 joules when it hits the ground, making option A. the correct option.

To answer this question, let's analyze the energy transformations involved. When the ball is thrown downward, it strikes the ground with 200 joules of kinetic energy.

This energy comes from the initial throw plus the gravitational potential energy converted during its fall.

When the second ball is thrown upward, it initially gains gravitational potential energy as it rises, then this potential energy is converted back to kinetic energy as it falls back down.

The initial speed given to the ball in both scenarios is the same, meaning the total energy input into the system is identical for both throws.

Since both balls have the same initial speed and both experience the same gravitational acceleration, the second ball will also have 200 joules of kinetic energy when it hits the ground.

Therefore, the correct answer is A. 200 J.

Answer:

Technician A

Explanation:

Often referred to as the profile or series, the aspect ratio of a tire is determined by dividing a tire’s section height by its section width when the tire is: inflated to maximum air pressure, mounted on the approved measuring rim, and under no load. This rules out Technician B.

A tire with a lower aspect ratio responds to lateral force more effectively than a tire with a higher aspect ratio. The aspect ratio affects steering stability. Generally, the shorter the sidewall, or the lower the aspect ratio, the less time it takes to transmit the steering input from the wheel to the tread. The result is quicker steering response. Aspect ratio also affects the tread contact patch. As a rule, a low-profile tire produces a wider tread contact patch. This wider tread contact patch creates a stiffer footprint that reduces distortion and provides improved cornering traction. Aspect ratio also impacts ride. A low-profile tire usually has a stiffer ride than the standard aspect ratio of 75 or more.

Final answer:

A. Technician A is correct about the aspect ratio of a tyre, while Technician B is incorrect about the impact of low aspect ratios on ride softness.

Explanation:

Technician A is correct. The aspect ratio of a tyre indeed represents the relationship between the tyre's cross-sectional height to its cross-sectional width. The aspect ratio is a critical factor in determining the tyre's performance and characteristics. Technician B is incorrect. Low aspect ratios do not provide a softer ride because they will deflect more over irregular surfaces.

In the given case, the aspect ratio primarily affects the tyre's handling, stability, and sidewall height. In conclusion, Technician A provides an accurate explanation regarding the aspect ratio of a tyre in relation to its dimensions and characteristics.

Explanation:

The relation between angular frequency and frequency is directly proportional:

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

So, if the angular frequency decreases, the frequency also decreases.

The relation between angular frequency and period is inversely proportional:

[tex]\omega=\frac{2\pi}{T}[/tex]

So, if the angular frequency decreases, the period increases.

When the angular frequency of an object in simple harmonic motion decreases, the period increases and the frequency decreases.

When the angular frequency of an object in simple harmonic motion decreases, the effect on the period and frequency is as follows:

Answer:

The depth of the crater is 20[m]

Explanation:

We can solve this problem using the kinematics equations, it should be noted that the negative sign of acceleration means that the drop of the object is down.

Data:

a = 1.6[m/s^2]

t = 5 [s]

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

v0= 0, because there is not initial veocity

y = final distance = 0 the bottom of the crater

g = a = acceleration of the crater [m/s^2]

y0= Point where the ball was dropped. = Distance from the original point to the bottom of the crater.

[tex]y_{0} =\frac{1}{2}*1.6*(5)^{2}\\  y_{0} = 20 [m][/tex]

Answer:31.62 m/s

Explanation:

Given

mass of body [tex]m=400 kg[/tex]

Pull on chain is [tex]F_1=6000g N=60 kN[/tex]

Pull get smaller at the rate of [tex]F_2=360g N/m[/tex]

Net Upward Force [tex]F=6000 g-360 g\times 10=24 kN[/tex]

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

[tex]a=\frac{24\times 1000}{m}[/tex]

[tex]a=\frac{24000}{400}[/tex]

[tex]a=60 m/s^2[/tex]

but g is acting downward

[tex]a_{net}=a-g=60-10=50 m/s^2[/tex]

using [tex]v^2-u^2=2 as[/tex]

here initial velocity is zero

[tex]v^2=2\times 50\times 10[/tex]

[tex]v=31.62 m/s[/tex]

The velocity of the body will be "31.62 m/s".

Given:

Mass of body,

Pull on chain,

The net upward force will be:

→ [tex]F = 6000-360\times 10[/tex]

      [tex]= 6000-3600[/tex]

      [tex]= 2400[/tex]

or,

      [tex]= 24 \ kN[/tex]

Now,

The net acceleration will be:

→ [tex]a = \frac{F}{m}[/tex]

     [tex]= \frac{24\times 1000}{400}[/tex]

     [tex]= 60 \ m/s^2[/tex]

But,

"g" is acting downwards then,

→ [tex]a_{net} = a -g[/tex]

          [tex]= 60-10[/tex]

          [tex]= 50 \ m/s^2[/tex]

By using,

→ [tex]v^2-u^2=2as[/tex]

          [tex]v^2= 2\times 50\times 10[/tex]

               [tex]= 1000[/tex]

            [tex]v = \sqrt{1000}[/tex]

               [tex]= 31.62 \ m/s[/tex]

Thus the above answer is right.

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Answer:

[tex]1.73553\times 10^{-10}\ m[/tex]

Explanation:

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

K = Potential difference = 50 V

m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]

The de broglie wavelength is given by

[tex]\lambda=\dfrac{h}{\sqrt{2mK}}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 50\times 1.6\times 10^{-19}}}\\\Rightarrow \lambda=1.73553\times 10^{-10}\ m[/tex]

The wavelength is [tex]1.73553\times 10^{-10}\ m[/tex]

Answer:

D.

Explanation:

8.1X10 to the power of 6 Joules.

Answer:

7.92×10^6Joules

Explanation:

Power is defined as the rate of change in work done. Mathematically, Power = Workdone/Time

Given power = 1.1×10³watts

Time = 2.0hours = 2×60×60

= 7,200seconds

Workdone = power × time

Work done = 1.1×10³×7200

Workdone = 7,920,000Joules

= 7.92×10^6Joules

Answer:

The speed of electron is [tex]8.7\times10^{6}\ m/s[/tex]

Explanation:

Given that,

Separation of the plate = 1.20 cm

Suppose the field is [tex]E=1.80\times10^{4}\ N/C[/tex].

If the electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate.

What the speed does it leave the hole?

We need to calculate the acceleration

Using formula of electric force

[tex]F = qE[/tex]

[tex]ma=qE[/tex]

[tex]a=\dfrac{qE}{m}[/tex]

We need to calculate the speed of electron

Using equation of motion

[tex]v^2=u^2+2as[/tex]

[tex]v^2=2as[/tex]

Put the value of acceleration in the formula

[tex]v^2=2\times\dfrac{qE}{m}\times s[/tex]

Put the value into the formula

[tex]v^2=2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}[/tex]

[tex]v=\sqrt{2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}}[/tex]

[tex]v=8.7\times10^{6}\ m/s[/tex]

Hence, The speed of electron is [tex]8.7\times10^{6}\ m/s[/tex]

Answer: True

Explanation: A knowledge engineer is a professional who works to improve and builds with the aid of science and also implements advanced logic and AI( artificial intelligence) in computer systems in order to investigate particular issues.

A knowledge engineer also helps tries to emulate the judgements of man.

Since you are part of the team involved in discovering errors in the electronic spread sheet you are a knowledge engineer.

Answer:

a = 2.5 m/s²

Explanation:

given,

initial speed = 0 m/s

final speed = 25 m/s

time = 10 s

acceleration of bike is equal to change in velocity per unit time

  [tex]a = \dfrac{\Delta v}{t}[/tex]

  [tex]a = \dfrac{v_f-v_i}{t}[/tex]

  [tex]a = \dfrac{25-0}{10}[/tex]

         a = 2.5 m/s²

hence, acceleration of bike is equal to a = 2.5 m/s²

The maximum allowable current in the cable for the experiment to be accurate, considering Earth's magnetic field is 0.5 Amps. This is calculated using the magnetic field strength formula and rearranging it to find the current.

To answer the question, we would need to use the formula to determine the magnetic field strength generated by a current carrying wire which is B = μ0 * I / 2πr, where B is the magnetic field strength generated by the wire, I is the current, r is the distance from the wire and `μ0` is the permeability of free space which is a constant equal to 4π * 10^-7 T m/A.

Let's rearrange this equation to make I the subject:

I = B * 2πr / μ0

We are told in the question that the maximum magnetic field (B) that does not adversely affect the experiment is 1.0 * 10^-4 T, and the distance (r) is 1.00 m. So substituting these numbers in:

I = 1.0 * 10^-4 T * 2π * 1.00 m / 4π * 10^-7 T m/A

This simplifies to 0.5 A. Therefore, the maximum allowable current in the wire is 0.5 Amps.

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Answer:

400-450 nm and 670-680 nm

Explanation:

Which wavelengths of light drive the highest rates of photosynthesis?

400-450 nm and 670-680 nm

Light in the violet-blue and red portions of the spectrum is most effective in driving photosynthesis

photosynthesis is the process through which green plant manufacture there food through sunlight, they synthesize nutrient in the this process through chlorophyll thereby releasing oxygen as a bye-product. a spectrum contains 7 colours combined . the violet-blue and red portions of the spectrum is most effective

Answer:

L= 1.06 m

Explanation:

frequency of a wave in a pipe closed at one end is given by

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

L= length of the pipe

v= velocity of wave

therefore, the length of the pipe

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

[tex]L= \frac{340}{4\times80}[/tex]

L= 1.06 m

L= 1.06 m far from the wall must you move to find the first quiet spot.

The distance from the wall(i.e the pipe length) you need to move to find the first quiet spot is 1.06 m.

The smallest standing wave pattern, i.e the first fundamental frequency or harmonic for a closed pipe usually consists of one node and an antinode. When the wavelength of a closed pipe of length L is 4 L, the basic standing wave is formed.

The frequency of a sound wave for a closed pipe at one end may thus be represented using the following formula:

[tex]\mathbf{f = \dfrac{v}{4L}}[/tex]

where;

Thus, the length of the pipe can be computed as:
[tex]\mathbf{L = \dfrac{340 \ m/s}{4\times 80 \ Hz}}[/tex]

L = 1.06 m

Learn more about the frequency of sound waves here:
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